I wrote about a way to generate more bases for classical logic about a year ago, when you have either CCpqCCqrCpr or CpCqp upfront in the basis. Examples of such systems are {CCpqCCqrCpr, CCNppp, CpCNpq}-Lukasiewicz, {CCpqCCqrCpr, CCpCpqCpq, CpCqp, CCNpNqCqp}-Tarski, {CpCqp, CCpCqrCCpqCpr, CCNpqCCNpNqp}-I don't think think I've seen this before, but it's basically implied by a system of Mendelsohn since if any system has [C] CCpCqrCqCpr provable in a proper sub-basis without axioms a, ..., z, of the basis, then a can get replaced by D[C].a, b can get replaced by D[C].b, ... z can get replaced by D[C].z. For example, for the basis {CCpCqrCCpqCpr, CpCqp, CCNpNqCqp}, [C] can get proved in the sub-basis {CpCqp, CCpCqrCCpqCpr}. Thus, CCNpNqCqp can get replaced by CpCCNqNpq for the basis {CCpCqrCCpqCpr, CpCqp, CCNpNqCqp}.
The following 4-basis for C-N classical logic is independent:
1. CCCCpqCrqsCCrps
2. CCpqCCNppq
3. CCCNpqrCpr
4. Cpp
Or as a system with fewer symbols, the axioms of the following are independent.
1. CCpqCCqrCpr
2. CCpqCCNppq
3. CpCNpq
4. Cpp
Does there exist a shorter 4-basis with independent axioms for C-N classical logic?
I note also that the following four formulas are not independent
1. CCpqCCqrCpr
2. CCNppp
3. CCCNpqrCpr
4. Cpp
since D3.2 yields Cpp.
Does there exist a 3-basis or a 2-basis for classical logic where one of the axioms is Cpp?
Friday, June 6, 2014
Wednesday, May 28, 2014
I ran Prover9 [1] with the following inputs:
1.
-P(i(x, y)) | -P(x) | P(y).
P(i(x,i(y,x))). %RVP
P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). %SD
2.
-P(i(x, y)) | -P(x) | P(y).
-P(i(x, y)) | -P(i(y, z)) | P(i(x, z)).
P(i(x,i(y,x))). %RVP
P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). %SD
-P(i(x, y)) | -P(i(y, z)) | P(i(x, z)) represents a derivable rule in the first system. I had the formula P(i(i(x, y), i(i(y, z), i(i(z, u), i(x, u))))) as the goal. The proof in the 1st system took 7.56 seconds, had 958 givens, generated 835656 formulas, and 23398 formulas kept. The proof in the 2nd system took 25.72 seconds, had 1450 givens, generated 3835597 formulas, and kept 26067 formulas. Both proofs took the same number of steps. The maximum clause weight of the second proof was lower than that of the first proof. As I understand things, this means that the longest formula of the second proof (excluding parentheses and predicate symbols) was shorter than that of the first proof. Here's the first proof:
% -------- Comments from original proof --------
% Proof 1 at 7.56 (+ 0.41) seconds.
% Length of proof is 21.
% Level of proof is 10.
% Maximum clause weight is 24.
% Given clauses 958.
1 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 P(i(x,i(y,x))). [assumption].
4 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
5 -P(i(i(c1,c2),i(i(c2,c3),i(i(c3,c4),i(c1,c4))))). [deny(1)].
6 P(i(x,i(y,i(z,y)))). [hyper(2,a,3,a,b,3,a)].
7 P(i(i(i(x,i(y,z)),i(x,y)),i(i(x,i(y,z)),i(x,z)))). [hyper(2,a,4,a,b,4,a)].
8 P(i(x,i(i(y,i(z,u)),i(i(y,z),i(y,u))))). [hyper(2,a,3,a,b,4,a)].
11 P(i(x,i(y,i(z,i(u,z))))). [hyper(2,a,3,a,b,6,a)].
23 P(i(i(x,i(i(y,x),z)),i(x,z))). [hyper(2,a,7,a,b,6,a)].
39 P(i(i(x,i(i(y,i(z,y)),u)),i(x,u))). [hyper(2,a,7,a,b,11,a)].
43 P(i(i(i(x,i(y,z)),i(i(i(x,y),i(x,z)),u)),i(i(x,i(y,z)),u))). [hyper(2,a,7,a,b,8,a)].
126 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(2,a,23,a,b,8,a)].
419 P(i(i(i(x,y),z),i(y,z))). [hyper(2,a,39,a,b,126,a)].
441 P(i(x,i(i(i(y,z),u),i(z,u)))). [hyper(2,a,3,a,b,419,a)].
876 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(2,a,43,a,b,441,a)].
1406 P(i(i(x,y),i(i(y,z),i(x,z)))). [hyper(2,a,876,a,b,126,a)].
2202 P(i(i(i(i(x,y),i(z,y)),u),i(i(z,x),u))). [hyper(2,a,1406,a,b,1406,a)].
2220 P(i(i(x,i(y,z)),i(x,i(i(z,u),i(y,u))))). [hyper(2,a,126,a,b,1406,a)].
23399 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))). [hyper(2,a,2202,a,b,2220,a)].
23400 $F. [resolve(23399,a,5,a)].
Here's the second proof:
% -------- Comments from original proof --------
% Proof 1 at 25.72 (+ 1.89) seconds.
% Length of proof is 21.
% Level of proof is 10.
% Maximum clause weight is 16.
% Given clauses 1450.
1 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 -P(i(x,y)) | -P(i(y,z)) | P(i(x,z)). [assumption].
4 P(i(x,i(y,x))). [assumption].
5 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
6 -P(i(i(c1,c2),i(i(c2,c3),i(i(c3,c4),i(c1,c4))))). [deny(1)].
8 P(i(x,i(y,i(z,y)))). [hyper(2,a,4,a,b,4,a)].
10 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(3,a,4,a,b,5,a)].
14 P(i(i(x,y),i(x,x))). [hyper(2,a,5,a,b,4,a)].
31 P(i(x,x)). [hyper(2,a,14,a,b,8,a)].
33 P(i(i(i(x,y),x),i(i(x,y),y))). [hyper(2,a,5,a,b,31,a)].
146 P(i(x,i(i(x,y),y))). [hyper(3,a,4,a,b,33,a)].
166 P(i(i(i(x,i(y,x)),z),z)). [hyper(2,a,33,a,b,8,a)].
190 P(i(i(x,y),i(x,i(i(y,z),z)))). [hyper(2,a,10,a,b,146,a)].
344 P(i(i(i(x,y),z),i(y,z))). [hyper(3,a,10,a,b,166,a)].
601 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(3,a,5,a,b,344,a)].
3542 P(i(i(x,y),i(i(y,z),i(x,z)))). [hyper(3,a,190,a,b,601,a)].
5634 P(i(i(x,y),i(i(z,i(y,u)),i(z,i(x,u))))). [hyper(3,a,3542,a,b,10,a)].
5639 P(i(i(i(i(x,y),i(z,y)),u),i(i(z,x),u))). [hyper(2,a,3542,a,b,3542,a)].
26068 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))). [hyper(3,a,5634,a,b,5639,a)].
26069 $F. [resolve(26068,a,6,a)].
The above may make it look like derived rules aren't of use. But I ran another trial looking for proofs of P(i(i(x, i(y, z)), i(y, i(x, z)))). For the first system I got the following proof:
% Proof 1 at 0.05 (+ 0.03) seconds.
% Length of proof is 17.
% Level of proof is 7.
% Maximum clause weight is 24.
% Given clauses 75.
1 P(i(i(x,i(y,z)),i(y,i(x,z)))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 P(i(x,i(y,x))). [assumption].
4 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
5 -P(i(i(c1,i(c2,c3)),i(c2,i(c1,c3)))). [deny(1)].
6 P(i(x,i(y,i(z,y)))). [hyper(2,a,3,a,b,3,a)].
7 P(i(i(i(x,i(y,z)),i(x,y)),i(i(x,i(y,z)),i(x,z)))). [hyper(2,a,4,a,b,4,a)].
8 P(i(x,i(i(y,i(z,u)),i(i(y,z),i(y,u))))). [hyper(2,a,3,a,b,4,a)].
11 P(i(x,i(y,i(z,i(u,z))))). [hyper(2,a,3,a,b,6,a)].
23 P(i(i(x,i(i(y,x),z)),i(x,z))). [hyper(2,a,7,a,b,6,a)].
39 P(i(i(x,i(i(y,i(z,y)),u)),i(x,u))). [hyper(2,a,7,a,b,11,a)].
43 P(i(i(i(x,i(y,z)),i(i(i(x,y),i(x,z)),u)),i(i(x,i(y,z)),u))). [hyper(2,a,7,a,b,8,a)].
126 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(2,a,23,a,b,8,a)].
419 P(i(i(i(x,y),z),i(y,z))). [hyper(2,a,39,a,b,126,a)].
441 P(i(x,i(i(i(y,z),u),i(z,u)))). [hyper(2,a,3,a,b,419,a)].
876 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(2,a,43,a,b,441,a)].
877 $F. [resolve(876,a,5,a)].
For the second system I got the following proof:
% -------- Comments from original proof --------
% Proof 1 at 0.03 (+ 0.01) seconds.
% Length of proof is 15.
% Level of proof is 7.
% Maximum clause weight is 14.
% Given clauses 38.
1 P(i(i(x,i(y,z)),i(y,i(x,z)))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 -P(i(x,y)) | -P(i(y,z)) | P(i(x,z)). [assumption].
4 P(i(x,i(y,x))). [assumption].
5 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
6 -P(i(i(c1,i(c2,c3)),i(c2,i(c1,c3)))). [deny(1)].
8 P(i(x,i(y,i(z,y)))). [hyper(2,a,4,a,b,4,a)].
10 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(3,a,4,a,b,5,a)].
14 P(i(i(x,y),i(x,x))). [hyper(2,a,5,a,b,4,a)].
31 P(i(x,x)). [hyper(2,a,14,a,b,8,a)].
33 P(i(i(i(x,y),x),i(i(x,y),y))). [hyper(2,a,5,a,b,31,a)].
166 P(i(i(i(x,i(y,x)),z),z)). [hyper(2,a,33,a,b,8,a)].
344 P(i(i(i(x,y),z),i(y,z))). [hyper(3,a,10,a,b,166,a)].
601 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(3,a,5,a,b,344,a)].
602 $F. [resolve(601,a,6,a)].
Since the second proof here took some 602 formulas in comparison to 877 formulas of the first proof, the second proof has length of 15 in comparison to length 17 of the first proof, the clause weight is 14 in comparison to 24, the given clauses is 38 in comparison to 75, and it took slightly less time this suggest that *sometimes* using derived rules might make it easier for Prover9 to find a proof.
[1] W. McCune, "Prover9 and Mace4", http://www.cs.unm.edu/~mccune/Prover9, 2005-2010.
1.
-P(i(x, y)) | -P(x) | P(y).
P(i(x,i(y,x))). %RVP
P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). %SD
2.
-P(i(x, y)) | -P(x) | P(y).
-P(i(x, y)) | -P(i(y, z)) | P(i(x, z)).
P(i(x,i(y,x))). %RVP
P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). %SD
-P(i(x, y)) | -P(i(y, z)) | P(i(x, z)) represents a derivable rule in the first system. I had the formula P(i(i(x, y), i(i(y, z), i(i(z, u), i(x, u))))) as the goal. The proof in the 1st system took 7.56 seconds, had 958 givens, generated 835656 formulas, and 23398 formulas kept. The proof in the 2nd system took 25.72 seconds, had 1450 givens, generated 3835597 formulas, and kept 26067 formulas. Both proofs took the same number of steps. The maximum clause weight of the second proof was lower than that of the first proof. As I understand things, this means that the longest formula of the second proof (excluding parentheses and predicate symbols) was shorter than that of the first proof. Here's the first proof:
% -------- Comments from original proof --------
% Proof 1 at 7.56 (+ 0.41) seconds.
% Length of proof is 21.
% Level of proof is 10.
% Maximum clause weight is 24.
% Given clauses 958.
1 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 P(i(x,i(y,x))). [assumption].
4 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
5 -P(i(i(c1,c2),i(i(c2,c3),i(i(c3,c4),i(c1,c4))))). [deny(1)].
6 P(i(x,i(y,i(z,y)))). [hyper(2,a,3,a,b,3,a)].
7 P(i(i(i(x,i(y,z)),i(x,y)),i(i(x,i(y,z)),i(x,z)))). [hyper(2,a,4,a,b,4,a)].
8 P(i(x,i(i(y,i(z,u)),i(i(y,z),i(y,u))))). [hyper(2,a,3,a,b,4,a)].
11 P(i(x,i(y,i(z,i(u,z))))). [hyper(2,a,3,a,b,6,a)].
23 P(i(i(x,i(i(y,x),z)),i(x,z))). [hyper(2,a,7,a,b,6,a)].
39 P(i(i(x,i(i(y,i(z,y)),u)),i(x,u))). [hyper(2,a,7,a,b,11,a)].
43 P(i(i(i(x,i(y,z)),i(i(i(x,y),i(x,z)),u)),i(i(x,i(y,z)),u))). [hyper(2,a,7,a,b,8,a)].
126 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(2,a,23,a,b,8,a)].
419 P(i(i(i(x,y),z),i(y,z))). [hyper(2,a,39,a,b,126,a)].
441 P(i(x,i(i(i(y,z),u),i(z,u)))). [hyper(2,a,3,a,b,419,a)].
876 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(2,a,43,a,b,441,a)].
1406 P(i(i(x,y),i(i(y,z),i(x,z)))). [hyper(2,a,876,a,b,126,a)].
2202 P(i(i(i(i(x,y),i(z,y)),u),i(i(z,x),u))). [hyper(2,a,1406,a,b,1406,a)].
2220 P(i(i(x,i(y,z)),i(x,i(i(z,u),i(y,u))))). [hyper(2,a,126,a,b,1406,a)].
23399 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))). [hyper(2,a,2202,a,b,2220,a)].
23400 $F. [resolve(23399,a,5,a)].
Here's the second proof:
% -------- Comments from original proof --------
% Proof 1 at 25.72 (+ 1.89) seconds.
% Length of proof is 21.
% Level of proof is 10.
% Maximum clause weight is 16.
% Given clauses 1450.
1 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 -P(i(x,y)) | -P(i(y,z)) | P(i(x,z)). [assumption].
4 P(i(x,i(y,x))). [assumption].
5 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
6 -P(i(i(c1,c2),i(i(c2,c3),i(i(c3,c4),i(c1,c4))))). [deny(1)].
8 P(i(x,i(y,i(z,y)))). [hyper(2,a,4,a,b,4,a)].
10 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(3,a,4,a,b,5,a)].
14 P(i(i(x,y),i(x,x))). [hyper(2,a,5,a,b,4,a)].
31 P(i(x,x)). [hyper(2,a,14,a,b,8,a)].
33 P(i(i(i(x,y),x),i(i(x,y),y))). [hyper(2,a,5,a,b,31,a)].
146 P(i(x,i(i(x,y),y))). [hyper(3,a,4,a,b,33,a)].
166 P(i(i(i(x,i(y,x)),z),z)). [hyper(2,a,33,a,b,8,a)].
190 P(i(i(x,y),i(x,i(i(y,z),z)))). [hyper(2,a,10,a,b,146,a)].
344 P(i(i(i(x,y),z),i(y,z))). [hyper(3,a,10,a,b,166,a)].
601 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(3,a,5,a,b,344,a)].
3542 P(i(i(x,y),i(i(y,z),i(x,z)))). [hyper(3,a,190,a,b,601,a)].
5634 P(i(i(x,y),i(i(z,i(y,u)),i(z,i(x,u))))). [hyper(3,a,3542,a,b,10,a)].
5639 P(i(i(i(i(x,y),i(z,y)),u),i(i(z,x),u))). [hyper(2,a,3542,a,b,3542,a)].
26068 P(i(i(x,y),i(i(y,z),i(i(z,u),i(x,u))))). [hyper(3,a,5634,a,b,5639,a)].
26069 $F. [resolve(26068,a,6,a)].
The above may make it look like derived rules aren't of use. But I ran another trial looking for proofs of P(i(i(x, i(y, z)), i(y, i(x, z)))). For the first system I got the following proof:
% Proof 1 at 0.05 (+ 0.03) seconds.
% Length of proof is 17.
% Level of proof is 7.
% Maximum clause weight is 24.
% Given clauses 75.
1 P(i(i(x,i(y,z)),i(y,i(x,z)))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 P(i(x,i(y,x))). [assumption].
4 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
5 -P(i(i(c1,i(c2,c3)),i(c2,i(c1,c3)))). [deny(1)].
6 P(i(x,i(y,i(z,y)))). [hyper(2,a,3,a,b,3,a)].
7 P(i(i(i(x,i(y,z)),i(x,y)),i(i(x,i(y,z)),i(x,z)))). [hyper(2,a,4,a,b,4,a)].
8 P(i(x,i(i(y,i(z,u)),i(i(y,z),i(y,u))))). [hyper(2,a,3,a,b,4,a)].
11 P(i(x,i(y,i(z,i(u,z))))). [hyper(2,a,3,a,b,6,a)].
23 P(i(i(x,i(i(y,x),z)),i(x,z))). [hyper(2,a,7,a,b,6,a)].
39 P(i(i(x,i(i(y,i(z,y)),u)),i(x,u))). [hyper(2,a,7,a,b,11,a)].
43 P(i(i(i(x,i(y,z)),i(i(i(x,y),i(x,z)),u)),i(i(x,i(y,z)),u))). [hyper(2,a,7,a,b,8,a)].
126 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(2,a,23,a,b,8,a)].
419 P(i(i(i(x,y),z),i(y,z))). [hyper(2,a,39,a,b,126,a)].
441 P(i(x,i(i(i(y,z),u),i(z,u)))). [hyper(2,a,3,a,b,419,a)].
876 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(2,a,43,a,b,441,a)].
877 $F. [resolve(876,a,5,a)].
For the second system I got the following proof:
% -------- Comments from original proof --------
% Proof 1 at 0.03 (+ 0.01) seconds.
% Length of proof is 15.
% Level of proof is 7.
% Maximum clause weight is 14.
% Given clauses 38.
1 P(i(i(x,i(y,z)),i(y,i(x,z)))) # label(non_clause) # label(goal). [goal].
2 -P(i(x,y)) | -P(x) | P(y). [assumption].
3 -P(i(x,y)) | -P(i(y,z)) | P(i(x,z)). [assumption].
4 P(i(x,i(y,x))). [assumption].
5 P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). [assumption].
6 -P(i(i(c1,i(c2,c3)),i(c2,i(c1,c3)))). [deny(1)].
8 P(i(x,i(y,i(z,y)))). [hyper(2,a,4,a,b,4,a)].
10 P(i(i(x,y),i(i(z,x),i(z,y)))). [hyper(3,a,4,a,b,5,a)].
14 P(i(i(x,y),i(x,x))). [hyper(2,a,5,a,b,4,a)].
31 P(i(x,x)). [hyper(2,a,14,a,b,8,a)].
33 P(i(i(i(x,y),x),i(i(x,y),y))). [hyper(2,a,5,a,b,31,a)].
166 P(i(i(i(x,i(y,x)),z),z)). [hyper(2,a,33,a,b,8,a)].
344 P(i(i(i(x,y),z),i(y,z))). [hyper(3,a,10,a,b,166,a)].
601 P(i(i(x,i(y,z)),i(y,i(x,z)))). [hyper(3,a,5,a,b,344,a)].
602 $F. [resolve(601,a,6,a)].
Since the second proof here took some 602 formulas in comparison to 877 formulas of the first proof, the second proof has length of 15 in comparison to length 17 of the first proof, the clause weight is 14 in comparison to 24, the given clauses is 38 in comparison to 75, and it took slightly less time this suggest that *sometimes* using derived rules might make it easier for Prover9 to find a proof.
[1] W. McCune, "Prover9 and Mace4", http://www.cs.unm.edu/~mccune/Prover9, 2005-2010.
Monday, May 19, 2014
Professor Ulrich asks "Can a fully-automated proof of
the sufficiency of Meredith's 21-character single axiom,
CCCCCpqCNrNsrtCCtpCsp
/
((((p⊃q)⊃(~r⊃~s))⊃r)⊃t)⊃((t⊃p)⊃(s⊃p)), for classical logic in C and
N be found that equals (or shortens) Meredith's own derivation of Syl
= CCpqCCqrCpr / ((p⊃q)⊃((q⊃r)⊃(p⊃r)),
Scotus = CpCNpq / p⊃(~p⊃q), and Clavius
= CCNppp / (~p⊃p)⊃p
using just 41 condensed detachments?
UPDATE: Larry Wos has (with OTTER) now discovered a proof using only 38 applications of condensed detachment. See Automated reasoning and the discovery of missing and elegant proofs, L. Wos and G. Peiper, Rinton, Paramus, 2003, sections 3.1 through 3.3."
The question seems local to the derivation of {CCpqCCqrCpr, CpCNpq, CCNppp}. One might ask "Can a fully-automated proof of the sufficiency of Meredith's 21-character single axiom, CCCCCpqCNrNsrtCCtpCsp / ((((p⊃q)⊃(~r⊃~s))⊃r)⊃t)⊃((t⊃p)⊃(s⊃p)), for classical logic in C and N be found that requires less than 38 condensed detachments to get to the axioms for any distinct basis of C-N classical logic?"
It takes no more than 5 condensed detachments to get to CpCqp.
UPDATE: Larry Wos has (with OTTER) now discovered a proof using only 38 applications of condensed detachment. See Automated reasoning and the discovery of missing and elegant proofs, L. Wos and G. Peiper, Rinton, Paramus, 2003, sections 3.1 through 3.3."
The question seems local to the derivation of {CCpqCCqrCpr, CpCNpq, CCNppp}. One might ask "Can a fully-automated proof of the sufficiency of Meredith's 21-character single axiom, CCCCCpqCNrNsrtCCtpCsp / ((((p⊃q)⊃(~r⊃~s))⊃r)⊃t)⊃((t⊃p)⊃(s⊃p)), for classical logic in C and N be found that requires less than 38 condensed detachments to get to the axioms for any distinct basis of C-N classical logic?"
It takes no more than 5 condensed detachments to get to CpCqp.
Tuesday, May 13, 2014
Wednesday, May 7, 2014
Formula of the day: CCCpqrCsCCqCrtCqt
The last two formulas of the day were CpCqp and CCpCqrCCpqCpr. Together these form a set of axioms sufficient for the positive implicational calculus (equivalently they form a "basis" for the positive implicational calculus). C. A. Meredith figured out that CCCpqrCsCCqCrtCqt itself serves as a basis for the positive implicational calculus of propositions. To prove that{CpCqp-RVP, CCpCqrCCpqCpr-SD} yields CCCpqrCsCCqCrtCqt we can construct a demonstration as follows, with Ci standing for conditional introduction:
hypothesis 1 ! CCpqr
hypothesis 2 !@ s
hypothesis 3 !@# CqCrt
hypothesis 4 !@#$ q
RVP 5 !@#$ CqCpq
D5.4 6 !@#$ Cpq
D1.6 7 !@#$ r
D3.4 8 !@#$ Crt
D8.7 9 !@#$ t
Ci 4-9 10 !@# Cqt
Ci 3-10 11 !@ CCqCrtCqt
Ci 2-11 12 ! CsCCqCrtCqt
Ci 1-12 13 CCCpqrCsCCqCrtCqt.
We can expand all of this out and write a formal proof of {CCCpqrCsCCqCrtCqt} from the basis {CpCqp-RVP, CCpCqrCCpqCpr-SD} using any decent proof of the deduction metatheorem. Here the letters a, b, ..., n, o can get substituted by something else, but p, q, ..., y, z cannot. First we'll take out the "$" lines.
hypothesis 1 ! CCpqr
hypothesis 2 !@ s
hypothesis 3 !@# CqCrt
D[RVP].1 4 !@# CaCCpqr This is C4-1
D[SD].4 5 !@# CCaCpqCar
D5.RVP 6 !@# Cqr ... C4-7
D[RVP].3 7 !@# CaCqCrt ... C4-3
D[SD].7 8 !@# CCaqCaCrt
D8.[Caa] 9 !@# CqCrt ... C4-8
D[SD].9 10 !@# CCqrCqt
D10.6 11 !@# Cqt C4-9
.
.
.
Now for the "#" lines in the above demonstration. C3-3 and C3-9 match exactly. C3-10 we can obtain simply from SD just by substitution.
hypothesis 1 ! CCpqr
hypothesis 2 !@ s
theorem 3 !@ CCaCbCcdCaCCbcCbd (we'll use this when we used "SD" above)
D[RVP].1 4 !@ CaCCpqr
D[RVP].4 5 !@ CaCbCCpqr ... C3-4
D3.5 6 !@ CaCCbCpqCbr ... C3-5
theorem 7 !@ CaCbCcb
D[SD].6 8 !@ CCaCbCpqCaCbr
D8.7 9 !@ CaCqr ... C3-6
D3.RVP 10 !@ CCabCCcaCcb ... C3-8
D[SD].SD 11 !@ CCCaCbcCabCCaCbcCac
D11.9 12 !@ CCqCraCqa C3-11
Now the "@" lines.
hypothesis 1 ! CCpqr
theorem 2 ! CCaCbCcdCaCCbcCbd
theorem 3 ! CCaCbCcCdeCaCbCCcdCce (we'll use this where we used 3 above)
D[RVP].1 4 ! CaCCpqr C2-1
D[RVP].4 5 ! CaCbCCpqr C2-4
D[RVP].5 6 ! CaCbCcCCpqr ... C2-5
D3.6 7 ! CaCbCCcCpqCcr ... C2-6
theorem 8 ! CaCbCcCdc ... C2-7
D2.7 9 ! CaCCbCcCpqCbCcr ... C2-8
D[SD].9 10 ! CCaCbCcCpqCaCbCcr
D10.8 11 ! CaCbCqr ... C2-9
theorem 12 ! CaCbCcb
D3.12 13 ! CaCCbcCCdbCdc ... C2-10
D[SD].SD 14 ! CCCaCbcCabCCaCbcCac
D[RVP].14 15 ! CaCCCbCcdCbcCCbCcdCbd ... C2-11
D[SD].15 16 ! CCaCCbCcdCbcCaCCbCcdCbd
D16.11 17 ! CaCCqCrbCqr ... C2-12
And now we'll write the formal proof (allowing substitution for all lower case letters):
axiom 1 CpCqp C1-4
axiom 2 CCpCqrCCpqCpr
D1.2 3 CpCCqCrsCCqrCqs
D2.3 4 CCpCqCrsCpCCqrCqs ... SD above
D1.4 5 CpCCqCrCstCqCCrsCrt
D2.5 6 CCpCqCrCstCpCqCCrsCrt 2 above
D1.6 7 CpCCqCrCsCtuCqCrCCstCsu
D2.7 8 CCpCqCrCsCtuCpCqCrCCstCsu ... 3 above
D1.1 9 CpCqCrq
D2.9 10 CCpqCpCrq
D10.1 11 CpCqCrp ... C1-5
D10.11 12 CpCqCrCsp ... C1-6
D1.9 13 CpCqCrCsr ... C1-12
D1.13 14 CpCqCrCsCts ... C1-8
D8.12 15 CCpqCrCsCCtpCtq ... C1-7
D6.15 16 CCpqCrCCsCtpCsCtq ... C1-9
D4.16 17 CCpqCCrCsCtpCrCsCtq ... C1-10
D2.17 18 CCCpqCrCsCtpCCpqCrCsCtq
D18.14 19 CCCpqrCsCtCqr ... C1-11
D8.13 20 CpCqCCrsCCtrCts ... C1-13
D4.3 21 CpCCCqCrsCqrCCqCrsCqs ... C1-14
D1.21 22 CpCqCCCrCstCrsCCrCstCrt ... C1-15
D4.22 23 CpCCqCCrCstCrsCqCCrCstCrt ... C1-16
D2.23 24 CCpCqCCrCstCrsCpCqCCrCstCrt
D24.19 25 CCCpqrCsCCqCrtCqt
I'll point out that CCpqCCrpCrq is a theorem of this system as follows:
axiom 1 CpCqp
axiom 2 CCpCqrCCpqCpr
D1.2 3 CpCCqCrsCCqrCqs
D2.3 4 CCpCqCrsCpCCqrCqs
D4.1 5 CCpqCCrpCrq
But, even though it did appear in the set-ups to prove CCCpqrCsCCqCrtCqt, we never produced an actual proof of CCpqCCrpCrq, though we did produce a proof of CpCqCCrsCCtrCts in which CCpqCrpCrq can get said to qualify as contained within. I've used Prover9 to find a proof of {CpCqp, CCpCqrCCpqCpr} from {CCCpqrCsCCqCrtCqt}. Meredith also found that CtCCpqCCCspCqrCpr can also serve as a basis of the positive implicational calculus. Ted Ulrich has found some other single axioms for this system here.
Both axioms, "CtCCpqCCCspCqrCpr" and "CCCpqrCsCCqCrtCqt" have 17 symbols. A stronger system, the implicational calculus of propositions, has a single axiom with just 13 symbols.
Also, the conversion procedure I used implicitly here depends on *how* a particular meta-proof of the deduction meta-theorem works. If we have a different meta-proof of the deduction meta-theorem, we'll have a different conversion procedure. My May 10th answer here shows the same process via a distinct meta-proof of the meta-deduction theorem.
The last two formulas of the day were CpCqp and CCpCqrCCpqCpr. Together these form a set of axioms sufficient for the positive implicational calculus (equivalently they form a "basis" for the positive implicational calculus). C. A. Meredith figured out that CCCpqrCsCCqCrtCqt itself serves as a basis for the positive implicational calculus of propositions. To prove that{CpCqp-RVP, CCpCqrCCpqCpr-SD} yields CCCpqrCsCCqCrtCqt we can construct a demonstration as follows, with Ci standing for conditional introduction:
hypothesis 1 ! CCpqr
hypothesis 2 !@ s
hypothesis 3 !@# CqCrt
hypothesis 4 !@#$ q
RVP 5 !@#$ CqCpq
D5.4 6 !@#$ Cpq
D1.6 7 !@#$ r
D3.4 8 !@#$ Crt
D8.7 9 !@#$ t
Ci 4-9 10 !@# Cqt
Ci 3-10 11 !@ CCqCrtCqt
Ci 2-11 12 ! CsCCqCrtCqt
Ci 1-12 13 CCCpqrCsCCqCrtCqt.
We can expand all of this out and write a formal proof of {CCCpqrCsCCqCrtCqt} from the basis {CpCqp-RVP, CCpCqrCCpqCpr-SD} using any decent proof of the deduction metatheorem. Here the letters a, b, ..., n, o can get substituted by something else, but p, q, ..., y, z cannot. First we'll take out the "$" lines.
hypothesis 1 ! CCpqr
hypothesis 2 !@ s
hypothesis 3 !@# CqCrt
D[RVP].1 4 !@# CaCCpqr This is C4-1
D[SD].4 5 !@# CCaCpqCar
D5.RVP 6 !@# Cqr ... C4-7
D[RVP].3 7 !@# CaCqCrt ... C4-3
D[SD].7 8 !@# CCaqCaCrt
D8.[Caa] 9 !@# CqCrt ... C4-8
D[SD].9 10 !@# CCqrCqt
D10.6 11 !@# Cqt C4-9
.
.
.
Now for the "#" lines in the above demonstration. C3-3 and C3-9 match exactly. C3-10 we can obtain simply from SD just by substitution.
hypothesis 1 ! CCpqr
hypothesis 2 !@ s
theorem 3 !@ CCaCbCcdCaCCbcCbd (we'll use this when we used "SD" above)
D[RVP].1 4 !@ CaCCpqr
D[RVP].4 5 !@ CaCbCCpqr ... C3-4
D3.5 6 !@ CaCCbCpqCbr ... C3-5
theorem 7 !@ CaCbCcb
D[SD].6 8 !@ CCaCbCpqCaCbr
D8.7 9 !@ CaCqr ... C3-6
D3.RVP 10 !@ CCabCCcaCcb ... C3-8
D[SD].SD 11 !@ CCCaCbcCabCCaCbcCac
D11.9 12 !@ CCqCraCqa C3-11
Now the "@" lines.
hypothesis 1 ! CCpqr
theorem 2 ! CCaCbCcdCaCCbcCbd
theorem 3 ! CCaCbCcCdeCaCbCCcdCce (we'll use this where we used 3 above)
D[RVP].1 4 ! CaCCpqr C2-1
D[RVP].4 5 ! CaCbCCpqr C2-4
D[RVP].5 6 ! CaCbCcCCpqr ... C2-5
D3.6 7 ! CaCbCCcCpqCcr ... C2-6
theorem 8 ! CaCbCcCdc ... C2-7
D2.7 9 ! CaCCbCcCpqCbCcr ... C2-8
D[SD].9 10 ! CCaCbCcCpqCaCbCcr
D10.8 11 ! CaCbCqr ... C2-9
theorem 12 ! CaCbCcb
D3.12 13 ! CaCCbcCCdbCdc ... C2-10
D[SD].SD 14 ! CCCaCbcCabCCaCbcCac
D[RVP].14 15 ! CaCCCbCcdCbcCCbCcdCbd ... C2-11
D[SD].15 16 ! CCaCCbCcdCbcCaCCbCcdCbd
D16.11 17 ! CaCCqCrbCqr ... C2-12
And now we'll write the formal proof (allowing substitution for all lower case letters):
axiom 1 CpCqp C1-4
axiom 2 CCpCqrCCpqCpr
D1.2 3 CpCCqCrsCCqrCqs
D2.3 4 CCpCqCrsCpCCqrCqs ... SD above
D1.4 5 CpCCqCrCstCqCCrsCrt
D2.5 6 CCpCqCrCstCpCqCCrsCrt 2 above
D1.6 7 CpCCqCrCsCtuCqCrCCstCsu
D2.7 8 CCpCqCrCsCtuCpCqCrCCstCsu ... 3 above
D1.1 9 CpCqCrq
D2.9 10 CCpqCpCrq
D10.1 11 CpCqCrp ... C1-5
D10.11 12 CpCqCrCsp ... C1-6
D1.9 13 CpCqCrCsr ... C1-12
D1.13 14 CpCqCrCsCts ... C1-8
D8.12 15 CCpqCrCsCCtpCtq ... C1-7
D6.15 16 CCpqCrCCsCtpCsCtq ... C1-9
D4.16 17 CCpqCCrCsCtpCrCsCtq ... C1-10
D2.17 18 CCCpqCrCsCtpCCpqCrCsCtq
D18.14 19 CCCpqrCsCtCqr ... C1-11
D8.13 20 CpCqCCrsCCtrCts ... C1-13
D4.3 21 CpCCCqCrsCqrCCqCrsCqs ... C1-14
D1.21 22 CpCqCCCrCstCrsCCrCstCrt ... C1-15
D4.22 23 CpCCqCCrCstCrsCqCCrCstCrt ... C1-16
D2.23 24 CCpCqCCrCstCrsCpCqCCrCstCrt
D24.19 25 CCCpqrCsCCqCrtCqt
I'll point out that CCpqCCrpCrq is a theorem of this system as follows:
axiom 1 CpCqp
axiom 2 CCpCqrCCpqCpr
D1.2 3 CpCCqCrsCCqrCqs
D2.3 4 CCpCqCrsCpCCqrCqs
D4.1 5 CCpqCCrpCrq
But, even though it did appear in the set-ups to prove CCCpqrCsCCqCrtCqt, we never produced an actual proof of CCpqCCrpCrq, though we did produce a proof of CpCqCCrsCCtrCts in which CCpqCrpCrq can get said to qualify as contained within. I've used Prover9 to find a proof of {CpCqp, CCpCqrCCpqCpr} from {CCCpqrCsCCqCrtCqt}. Meredith also found that CtCCpqCCCspCqrCpr can also serve as a basis of the positive implicational calculus. Ted Ulrich has found some other single axioms for this system here.
Both axioms, "CtCCpqCCCspCqrCpr" and "CCCpqrCsCCqCrtCqt" have 17 symbols. A stronger system, the implicational calculus of propositions, has a single axiom with just 13 symbols.
Also, the conversion procedure I used implicitly here depends on *how* a particular meta-proof of the deduction meta-theorem works. If we have a different meta-proof of the deduction meta-theorem, we'll have a different conversion procedure. My May 10th answer here shows the same process via a distinct meta-proof of the meta-deduction theorem.
Tuesday, May 6, 2014
Logical Formula of the Day: self-distribution CCpCqrCCpqCpr.
Here's two links on condensed detachment. If we have two formulas Cab, and "a", condensed detachment produces the most general formula "b". For instance, if we have
1. CpCqp as "a" and
2. CCpCqrCCpqCpr as "Cab",
we could substitute "r" in CCpCqrCCpqCpr, and thus obtain CCpqCpp from a condensed detachment D2.1. We could also substitute "r" with "Cqp", and "p" with "Cqp" in CCpCqrCCpqCpr with Cqp obtaining CCCqpCqCqpCCCqppCCqpCqp, and then substitute "p" with "Cqp" in CpCqp obtaining CCqpCqCqp. Thus, from 1. and 2. we could detach CCCqpqCCqpCqp as a theorem. However, CCCqpqCCqpCqp is less general than CCpqCpp, since if we substitute "p" with "Cqp" in CCpqCpp, then we obtain CCCqpqCCqpCqp. So, condensed detachment Da.b obtains a form from which you can obtain every other theorem detachable from a substitution instance of "a" as the major premise and "b" as the minor premise.
Let's look a little at the system {CCpCqrCCpqCpr}
axiom 1 CCpCqrCCpqCpr
D1.1 2 CCCpCqrCpqCCpCqrCpr
D1.2 3 CCCCpCqrCpqCpCqrCCCpCqrCpqCpr
D1.3 4 CCCCCpCqrCpqCpCqrCCpCqrCpqCCCCpCqrCpqCpCqrCpr
Every theorem of this system obtained via condensed detachment has three variables. This follows, because the axiom has only three variables in the antecedent and the consequent. Thus, the second theorem has only three variables in the antecedent and the consequent.
If we have CCpCqrCCpqCpr and CpCqp as axioms or theorems, then we have a deduction metatheorem for the system which says "if we have an assumption "a" which leads to a formula "b" under the scope of some set of assumptions "g", then Cab holds under the scope of the assumptions "g" also". Or more concisely we can write:
Deduction Metatheorem: If {g, a} |- b, then {g}|-Cab.
Here's two links on condensed detachment. If we have two formulas Cab, and "a", condensed detachment produces the most general formula "b". For instance, if we have
1. CpCqp as "a" and
2. CCpCqrCCpqCpr as "Cab",
we could substitute "r" in CCpCqrCCpqCpr, and thus obtain CCpqCpp from a condensed detachment D2.1. We could also substitute "r" with "Cqp", and "p" with "Cqp" in CCpCqrCCpqCpr with Cqp obtaining CCCqpCqCqpCCCqppCCqpCqp, and then substitute "p" with "Cqp" in CpCqp obtaining CCqpCqCqp. Thus, from 1. and 2. we could detach CCCqpqCCqpCqp as a theorem. However, CCCqpqCCqpCqp is less general than CCpqCpp, since if we substitute "p" with "Cqp" in CCpqCpp, then we obtain CCCqpqCCqpCqp. So, condensed detachment Da.b obtains a form from which you can obtain every other theorem detachable from a substitution instance of "a" as the major premise and "b" as the minor premise.
Let's look a little at the system {CCpCqrCCpqCpr}
axiom 1 CCpCqrCCpqCpr
D1.1 2 CCCpCqrCpqCCpCqrCpr
D1.2 3 CCCCpCqrCpqCpCqrCCCpCqrCpqCpr
D1.3 4 CCCCCpCqrCpqCpCqrCCpCqrCpqCCCCpCqrCpqCpCqrCpr
Every theorem of this system obtained via condensed detachment has three variables. This follows, because the axiom has only three variables in the antecedent and the consequent. Thus, the second theorem has only three variables in the antecedent and the consequent.
If we have CCpCqrCCpqCpr and CpCqp as axioms or theorems, then we have a deduction metatheorem for the system which says "if we have an assumption "a" which leads to a formula "b" under the scope of some set of assumptions "g", then Cab holds under the scope of the assumptions "g" also". Or more concisely we can write:
Deduction Metatheorem: If {g, a} |- b, then {g}|-Cab.
Monday, May 5, 2014
Formula of the Day: Recursive Variable Prefixing CpCqp.
When proving that a logical system has a deduction metatheorem, CpCqp proves indispensable. In fact, CpCqp along with CCpCqrCCpqCpr or CCpqCCpCqrCpr suffices to prove a deduction metatheorem. Let's consider the system {CpCqp}. For this system:
Lemma: Every theorem obtainable by condensed detachment for the system {CpCqp} has the form C0C1...CnCpCqp.
Demonstration: The first theorem obtainable by condensed detachment is CpCqCrq. Now from {CpCqp, CpCqCrq} we can use either thesis as the minor premise, and either one as the major. If CpCqp is the major, then we will detach Cab where "a" is a variable and "b" consists of the minor premise. If CpCqCrq is the major premise, then we will obtain CpCqp. More generally, if we have CpCqp as the major premise and use any theorem "b" as the minor premise, we will obtain Cab upon a condensed detachment. If Cat, where "a" does not appear in "t" gets used as the major premise, then we will obtain "t" after a condensed detachment. Thus, any time we use condensed detachment in this system we'll either obtain CpCqp or some theorem which has CpCqp as its last 5 letters.
The above lemma can get argued also via mathematical induction.
Corollary: Every thesis (theorem or axiom) obtainable via condensed detachment for the system {CpCqp} can serve as a sole axiom for a system with the same theses as {CpCqp}.
Demonstration: Since every axiom has form C0C1...CnCpCqp by the above lemma, we can obtain CpCqp as a theorem of the system. Since CpCqp has C0C1...CnCpCqp as a theorem, it follows that the systems have the exact same theorems.
Consequently it follows that in addition to {CpCqp, CCpCqrCCpqCpr} sufficing to prove the deduction theorem *any* theorem obtainable from the system {CpCqp} solely by condensed detachment can take the place of CpCqp in {CpCqp, CCpCqrCCpqCpr} .
When proving that a logical system has a deduction metatheorem, CpCqp proves indispensable. In fact, CpCqp along with CCpCqrCCpqCpr or CCpqCCpCqrCpr suffices to prove a deduction metatheorem. Let's consider the system {CpCqp}. For this system:
Lemma: Every theorem obtainable by condensed detachment for the system {CpCqp} has the form C0C1...CnCpCqp.
Demonstration: The first theorem obtainable by condensed detachment is CpCqCrq. Now from {CpCqp, CpCqCrq} we can use either thesis as the minor premise, and either one as the major. If CpCqp is the major, then we will detach Cab where "a" is a variable and "b" consists of the minor premise. If CpCqCrq is the major premise, then we will obtain CpCqp. More generally, if we have CpCqp as the major premise and use any theorem "b" as the minor premise, we will obtain Cab upon a condensed detachment. If Cat, where "a" does not appear in "t" gets used as the major premise, then we will obtain "t" after a condensed detachment. Thus, any time we use condensed detachment in this system we'll either obtain CpCqp or some theorem which has CpCqp as its last 5 letters.
The above lemma can get argued also via mathematical induction.
Corollary: Every thesis (theorem or axiom) obtainable via condensed detachment for the system {CpCqp} can serve as a sole axiom for a system with the same theses as {CpCqp}.
Demonstration: Since every axiom has form C0C1...CnCpCqp by the above lemma, we can obtain CpCqp as a theorem of the system. Since CpCqp has C0C1...CnCpCqp as a theorem, it follows that the systems have the exact same theorems.
Consequently it follows that in addition to {CpCqp, CCpCqrCCpqCpr} sufficing to prove the deduction theorem *any* theorem obtainable from the system {CpCqp} solely by condensed detachment can take the place of CpCqp in {CpCqp, CCpCqrCCpqCpr} .
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