Mimic Operations

The author writes informal proofs in the following.  The author uses Reverse Polish Notation also.

Suppose we have an infinite set S, and a operation "*". A mimic operation *' of * consists of an operation on S such that for all but a finite number of points a₁, ..., a_n *=a₁, ..., a_n *'. So, for a unary operation F on the natural numbers N, a mimic operation F' satisfies xF=xF' except for a finite set of natural numbers {a₁,...,a_n}. For a binary operation G on N, a mimic of B, B' satisfies xyB=xyB' except for a finite number of pairs of natural numbers {(a, b)₁,...,(a, b)_n} and so on.

Theorem 1: For any n-ary operation O on an infinite set, there exists an infinity of mimic operations, O'.

Proof: An n-ary operation O can get defined by a set of n 1 + tuples. E. G. the binary operation of addition can get defined by the triples (2, 4, 6), (1, 1, 2) such that for (x, y, z), xy+=z. Now consider the set of such tuples T for O. Vary the n 1 + part of the tuples in T for some finite number of tuples. This forms a set of tuples T' which describes an n-ary operation distinct from O in but a finite number of points, and thus forms a mimic operation O'. Note that since there exist an infinity of tuples belonging to T, there exists an infinity of ways to form T' given T. Thus, given an n-ary operation O, there exists an infinity of mimic operations O'.

Define a k-mimic operation M^k of an operation M as an operation which differs from M by but k points, where both operations operate over an infinite set. In other words, for an operation M_n of arity n with mimic M^k_n, a₁...a_nM_n=a₁...a_nM^k_n, except for k tuples {(b₁, ..., b_n), ..., (l₁, ..., l_n)} where this equality fails.

Theorem 2: For any n-ary operation O on an infinite set, there exists an infinity of k mimic operations.

Proof: Since given the set of tuples T for an operation O there exist an infinity of ways to form T' (see the proof of Theorem 1). From T' we have an infinity of k-mimic operations O' of O.

Theorem 3: If a binary operation O satisfies commutativity, then if xxO=xxO¹,  O¹ does not satisfy commutativity.

Proof: If O¹ is a 1-mimic of O, then xyO=xyO¹ for all but one pair (a, b). Suppose that for all x and y, if xyO=yxO, and xyO¹=yxO¹.  Suppose that (a, b) consists of the point where O and O¹ differ, by say letting abO=c, and letting abO¹=d, where d does not equal c.  It follows then that abO¹=d=baO¹ by supposition of commutation for O¹. But, this implies that the pair (b, a) consists of a second point where O and O¹ differ, which contradicts the supposition of O¹ as a 1-mimic of O.  Consequently, the theorem follows and if O¹ satisfies commutation it at least consists of a 2-mimic.

Conjecture: If Oⁿ and O^k are mimics of operation O, then Oⁿ and O^k are mimics of each other.

The author thinks theorem 3 generalizes to o-mimics, where "o" indicates any positive odd number. The author also thinks that converse of the theorem holds in that if a 1-mimic does satisfy commutation, the operation which it mimics will not satisfy commutation also.  Do there exist any relationships (or lack thereof) between an associative operation and its mimics?  Between an idempotent operation and its mimics?

Alternative Ways to Approach Propositional Calculus

There exist several ways to approach propositional calculus: axiomatically, through natural deduction, sequent calculi, semantic tableaux. An axiomatic presentation sometimes will start with a list of axioms, the rule of substitution, and the rule of conditional elimination (Co for "conditional out"). For example, one could start with the axiom set {CCpqCCqrCpr, CCNppp, CpCNpq} A1 or {CqCpq, CCpCqrCCpqCpr, CCNpNqCqp} A2. But, let's say we start another way.

Suppose we have the rules of conditional proof alon with substitution and conditional elimination. In other words, we have a rule which allows to introduce any propositional form as a hypothesis, so long as we keep track of its scope, and we have a rule of conditional introduction (Ci for "conditional in"). Suppose also, we don't want to use any axioms for propositional calculus. What sorts of rules of inference do we need?

Well, since the above axiom sets work as sufficient for propositional calculus, we could use the deduction metatheorem from CCNppp to obtain the rule of inference CNpp|-p. So, instead of using A1 we could use R3 {(Cpq, Cqr, p |-r), (CNpp|-p), (p, Np|-q)} along with just conditional elimination and substitution for propositional calculus. But, really, given the rules of conditional proof, we don't need all those rules as primitive rules. (Cpq, Cqr, p |-r) can get derived using just conditional proof as follows:

1 Cpq hypothesis
2 Cqr hypothesis
3 p hypothesis
4 q 3, 1 Co
5 r 4, 2 Co

We can also derive CqCpq, and CCpCqrCCpqCpr and using just conditional proof as follows:

1 | q hypothesis
2 || p hypothesis
3 ||| q hypothesis
4 || Cqq 3-3 Ci
5 || q 1, 4 Co
6 | Cpq 2-5 Ci
7 CqCpq 1-6 Ci

1 | CpCqr hypothesis
2 || Cpq hypothesis
3 ||| p hypothesis
4 ||| q 3, 2 Co
5 ||| Cqr 3, 1 Co
6 ||| r 4, 5 Co
7 || Cpr 3-6 Ci
8 | CCpqCpr 2-7 Ci
9 CCpCqrCCpqCpr 1-8 Ci

So, by the first proof instead of considering A1 with substitution and conditional elimination, we can consider {CCNppp, CpCNpq} N1 along with the rules of conditional proof, conditional elimination and substitution. By the second and third proofs instead of considering A2 we can consider N2 {CCNpNqCqp} along with the same rules which go along with N1. Now using the deduction metatheorem instead of considering N1 we can consider R1 {(CNpp|-p) R1a, (p, Np|-q) R1b} and R2 {(CNpNq, q |-p) MT}. A few weeks ago I took an old logic book and tried to prove certain theorems in both R1 and R2.

So, let's compare some proofs of Peirce's law CCCpqpp in R2 and R1 (of course more proofs exist, the following just give examples). In R1 a proof might go:

1 | CCpqp hypothesis
2 || Np hypothesis
3 ||| p hypothesis
4 ||| q 2, 3 R1b
5 || Cpq 3-4 Ci
6 || p 5, 1 Co
7 | CNpp 2-6 Ci
8 | p R1a
9 CCCpqpp 1-8 Ci

In R2 a proof might go:

1 | CCpqp hypothesis
2 || Np hypothesis
3 ||| p hypothesis
4 |||| Nq hypothesis
5 ||||| Np hypothesis
6 |||| CNpNp 5-5 Ci
7 |||| Np 2, 6 Co
8 ||| CNqNp 4-7 Ci
9 ||| q 3, 8 MT
10 || Cpq 2-9 Ci
11 || p 10, 1, Co
12 | CNpp 2-11 Ci
13 || Np hypothsis
14 ||| NNCNpp hypothesis
15 |||| NNNp hypothesis
16 ||||| Np hypothesis
17 |||| CNpNp 16-16 Ci
18 |||| Np 13, 17 Co
19 ||| CNNNpNp 15-18 Ci
20 ||| p 13, 12 Co
21 ||| NNp 20, 19 MT
22 || CNNCNppNNp 14-21 Ci
23 || NCNpp 13, 22 MT
24 | CNpNCNpp 13-23 Ci
25 | p 12, 24 MT
26 CCCpqpp 1-25 Ci

Let's also compare proofs of CNNpp and CpNNp. In R1 a proof for CNNpp might go:

1 | NNp hypothesis
2 || Np hypothesis
3 || p 1, 2 R1b
4 | CNpp 2-3 Ci
5 | p 4, R1a
6 CNNpp 1-5 Ci

In R2 a proof for CNNpp might go:

1 | NNp hypothesis
2 || Np hypothesis
3 ||| NNNNp hypothesis
4 |||| NNp hypothesis
5 ||| CNNpNNp 4-4 Ci
6 ||| NNp 1, 5 Co
7 || CNNNNpNNp 3-6 Ci
8 || NNNp 2, 7 MT
9 | CNpNNNp 2-8 Ci
10 | p 1, 9 MT
11 CNNpp 1-10 Ci.

For CpNNp in R1 we can have:

1 | p hypothesis
2 || Np hyopthesis
3 || NNNp 1, 2 R1b
4 | CNpNNNp 2-3 Ci
5 | NNp 4, R2b
6 CpNNp 1-5 Ci

For CpNNp in R2 we can have:

1 | p hypothesis
2 || NNNp hypothesis
3 ||| NNp hypothesis
4 |||| NNNNNp hypothesis
5 ||||| NNNp hypothesis
6 |||| CNNNpNNNp 5-5 Ci
7 |||| NNNp 2, 6 Co
8 ||| CNNNNNpNNNp 4-7 Ci
9 ||| NNNNp 3, 8 MT
10 || CNNpNNNNp 3-9 Ci
11 || Np 2, 10 MT
12 | CNNNpNp 2-11 Ci
13 | NNp 1, 12 MT
14 CpNNp 1-13 Ci.

Expressing All Statement Forms of Two Logical Variables as Conditionals

Every statement form which has only two logical variables can get expressed (i. e. comes as logically equivalent to) with {C, N} as the only set allowed for logical operations. For example, the statement form for conjunction Kxy, can get expressed as NCxNqy. Allowing only C and N as the only logical operations in a formula, can we express all statement forms with only two logical variables as conditionals? In other words, can we express all statement forms with only two logical variables such that C is the primary connective, or equivalently such that in Polish notation C is the first symbol of the statement form?

I answer yes. To see that this holds, you can consider the Sheffer stroke D:D00=1, D01=1, D10=1, D11=0. Since Nx==Dxx, and Dxy==CxNy (in other words we can define N and D in this way, and Nx comes as logically equivalent to Dxx, and Dxy comes as logically equivalent to CxNy), for any statement form which comes as a negation, we can transform it into a conditional, by first transforming it into a statement form with the Sheffer stroke D as its primary connective, and then transform the obtained statement form directly into a conditional.

More concretely one can see that you can do such as follows: let abcd indicate that the statement form has value "a" for the pair (0, 0) where 0 indicates falsity, and 1 truth, "b" for the pair (0, 1), c for the pair (1, 0), and d for the pair (1, 1). So, for example 0101 indicates that the statement form has truth value of false (0) when both arguments are false, has truth value of truth (1) when the first argument is false and the second true, has truth value of false (0) when the first argument is true and the second false, and has truth value of true (1) when both arguments are true. Thus, to show that all statement forms of two variables can get expressed as a conditional we only need to come with one example for 0000 through 1111.

For 0000, CCppNCpp works.
For 0001, from Kpq one can infer NCpNq, from which one can infer DCpNqCpNq, from which it follows that CCpNqNCpNq.
For 0010, from KpNq one can infer NCpq, to DCpqCpq to CCpqNCpq.
For 0011, from p we can infer CCqqp (in general C1y==y).
For 0100, from KqNp to NCqp, DCqpCqp, CCqpNCqp.
For 0101, from q to CCppq.
For 0110, from NEpq to NKCpqCqp to NNCCpqNCqp to CCpqNCqp
For 0111, Apq to CNpq
For 1000, KNpNq to NCNpNNq to NCNpq to DCNpqCNpq to CCNpqNCNpq
For 1001, from Epq to KCpqCqp to NCCpqNCqp to DCCpqNCqpCCpqNCqp to CCCpqNCqpNCCpqNCqp
For 1010, from Nq to CCppNq
For 1011, from NKNpq to NNCNpNq to CNpNq or Cqp
For 1100, from Np to CCqqNp
For 1101, Cpq
For 1110, NKpq to NNCpNq to CpNq
For 1111, Cpp.

With Conditional Proof You Never Need a Rule of Repetition

Suppose we have the rule of hypothesis introduction, conditional introduction, and conditional elimination. From just these rules of inference, in classical propositional calculus, we never need a rule of repetition, nor need CqCpq as an axiom. In other words, if those rules hold, then "q, p|-q" comes as a valid schema and CqCpq is a theorem.

1 | q hypothesis introduction
2 || p hypothesis introduction
3 ||| q hypothesis introduction
4 || Cqq 3-3 conditional introduction
5 || q 1, 4 conditional elimination
6 | Cpq 2-5 conditional introduction
7 CqCpq 1-6 conditional introduction.

Define a propositional form as follows:

1) Lower case letters, along with lower case letters subscripted with numbers a₁, b₂, c₁₃ are propositional forms.

2) If x and y are both propositional forms, then Cxy is a propositional form.

Let Cab denote the material conditional with 'a' and 'b' taking truth values in {T, F}. Let "C₁ ...C_n" denote a concatenation of n conditionals, and "a₁...a_(n+1)" a concatenation of (n+1) logical variables. E. G. C₁...C₃a₁...a₄ denotes CCCa₁a₂a₃a₄. Let a/b denote a substitution of variable "a" with "b", let a/Cab denote that we substitute variable "a" with Cab.

Axiom 1: CCpqCCqrCpr

Axiom 2: CqCpq

Theorem 1: CCCpqC₁...C_(10^100)a₁...a_((10^100)+1)CqC₁...C_(10^100)a₁...a_((10^100)+1)

Demonstration:

1 CCqCpqCCCpqC₁...C_(10^100)a₁...a_((10^100)+1)CqC₁...C_(10^100)
a₁...a_((10^100)+1) Axiom 1, p/q, q/Cpq, r/C₁...C_(10^100)a₁...a_((10^100)+1)

2 CCCpqC₁...C_(10^100)a₁...a_((10^100)+1)CqC₁...C_(10^100)a₁...a_((10^100)+1) Axiom 2, 1 detachment.

Is ApNp a Conjunctive Normal Form?

The notion of a conjunctive normal form first appears in Paul Bernays's second Habilitationsschrift [1] (doctoral thesis). In that text, Bernays established the completeness of propositional calculus [2]. On p. 38-39 of Formal Logic [3] Arthur Prior writes in the context of a completeness proof for propositional calculus:

"What they show [Hilbert and Ackermann] is that any formula of the propositional calculus is equivalent either (i) to a single propositional variable or the negation of one, or (ii) to an alternation of which all the members are either propositional variables or their negations, or alternations (or alternations of alternations, etc.) of such or (iii) to a conjunction of which all members are of the form listed under (i) and (ii), or conjunctions (or conjunctions of conjunctions, etc.) of such. We may describe all these forms as 'conjunctions of alternations of propositional-variables-or-their-negations', if (a) we call a form like 'KKpqr' a 'conjunction of p, q, and r' instead of calling it a 'conjunction of the-conjunction-of-p-and-q and r', and similarly with alternations, and (b) we regard a single propositional variable (or negation of one) as an alternation with a single alternant (it is as it were a limiting case of alternation; in any case such a variable is equivalent to the alternation of itself with itself-EpApp), and (c) we also regard a single propositional variable (or negation of one) as a conjunction with a single conjunct, and a single alternation as a conjunction-of-alternations with a single conjunct. The forms which would be 'conjunctions of alternations of propositional variables or their negations' would include such examples as

(i) p and Np (conjunctions-with-one-member of alternations-with-one-member).

(ii) ApNq (a conjunction-with-one-member of an alternation with two membes).

(iii) AApNqNp (a conjunction-with-one-member of an alternation with three members).

(iv) KpApq (a conjunction of an alternation-with-one-member and an alternation with two members).

(v) KApqApNq (a conjunction of two alternations-with-two-members).

(vi) KApAqrKApqApAst (a conjunction-with-three-members of an alternation-with-three-members, and an alternation-with-two-members, and another alternation-with-three-members.)

That every truth-functional formula can be 'put into normal form', i.e. shown to be equivalent to (i. e. to imply and be implied by) a formula of the above sort..."

So, for Prior, conditions (i), (ii), and (iii) in the first paragraph almost surely would define a "conjunctive normal form". This suggests that in the context of a (correct) proof of the completeness of propositional logic, that is, where the notion of a conjunctive normal form first got defined, ApNp is indeed a conjunctive normal form.

1 This information I retrieved http://books.google.com/books?id=1xEVkzuX5e0C&pg=PA501&lpg=PA501&dq=conjunctive+normal+form+bernays+Habilitationsschrift&source=bl&ots=DfCwTAO-wL&sig=rMok5p3XCMoGxQY8j-YEWLKCkes&hl=en&ei=GlZ2TpjNMoS6tgfu_6DSDA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBsQ6AEwAA#v=onepage&q=conjunctive%20normal%20form%20bernays%20Habilitationsschrift&f=false on September 18, 2011.

2 This information I retrieved from http://www.encyclopedia.com/doc/1G2-2830905020.html on September 18, 2011.

3 Prior, Arthur. Formal Logic Oxford University Press 1962.

Order AND Parentheses Free Notation?

In mathematics and logic expressions often get written with the operation in infix notation... we write a+b, a^b, a v b. As logicians and mathematicians know this means that we either have to establish an order of operations when we have more than operation involved, such as Exponents, Multiplication, Division, Addition, Subtraction for arithmetic, or use parentheses, or have ambiguous expressions. Since at least the time of Lukasiewicz, mathematical and logical scholars have known that we can stop excusing our dear Aunt Sally. In Lukasiewicz notation instead of writing (5+2)*4, we can simply write *+5 2 4. In reverse Lukasiewicz notation we can write 5 2+4*. I highly recommend that you read the link there, and translate some common logical or mathematical expressions into reverse Lukasiewicz notation. Since these notational systems establish that we can write mathematical and logical expressions in prefix notation (Lukasiewicz notation), infix notation, and postfix notation (reverse Lukasiewicz notation), can we write mathematical and logical expressions in a way such that the operations involved can come *anywhere* in the expression *without* having a fixed order as declined languages do?

For variables or constants, suppose we write a unique subscript below each of them. For an n-ary operation suppose we write n sequenced subscripts which correspond to the subscripts in the order of variables or constants it operates on. E. G. if we write x_a y_b+_ab, then we indicate that the operation "+" operates on x and y. We can then immediately write x_a+_aby_b as well as +_abx_ay_b. So, suppose we have a function F such that a b cF=b c*a+. Then we could write 4_c8_dF_cba6_a7_e+_de2_b. So, do you believe that this gives us an order free notation for the operations in a logical or mathematical expression? Please answer this question and write a response down on paper before reading my further comment.