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Larry Fiddick <lfidd…@sfu.ca> wrote:
>suppose you wanted to prove the following argument using the method of truth
>tables:

>[(P v Q) -> (Q v R)] & (P -> ~R)
>~Q
>————————————
>Therefore, ~P

>is there anything intrinsically wrong with proceeding as follows:

>1) [(P v Q) -> (Q v R)] & (P -> ~R) is defined as follows:

>P   Q       R       ||      [(P v Q) -> (Q v R)] & (P -> ~R)              
>————————————————————————-
>T   T       T       ||                      F      
>T   T       F       ||                      T      
>T   F       T       ||                      F
>T   F       F       ||                      F
>F   T       T       ||                      F
>F   T       F       ||                      T
>F   F       T       ||                      T
>F   F       F       ||                      T

>(ok, in order for things to fit, let’s call the formula ‘Z’)

>2) the truth table for the argument is:

>P   Q       R       ||      ~Q      Z       ~P
>———————————————————
>T   T       T       ||       F      F        F
>T   T       F       ||       F      T        F
>T   F       T       ||       T      F        F
>T   F       F       ||       T      F        F
>F   T       T       ||       F      F        T
>F   T       F       ||       F      T        T
>F   F       T       ||       T      T        T
>F   F       F       ||       T      T        T

>3) an inspection of the truth table shows that there are no instances in
>which ~Q and Z are true, and ~P is false, therefore the argument is valid.

        Right.  Every t.a. that satisfies {~Q, Z} also satisfies ~P;
that is,  {~Q, Z} |= ~P.

>So, in other words, does the method of truth tables permit one in principle to
>take the truth table definition of any complex formula, e.g. Z, as given, and
>then plug it in to the truth table to look for counterexamples?

>Now if this is permitted, if one can take the truth-table definition of any
>logical sentence as given, would this amount to a purely semantic deduction
>procedure? Suppose, instead, that one decomposed Z into its constituents in
>order to make the inference more transparent, would this decomposition be
>considered syntactic? (even though the search conducted after decomposing the
>sentence might again be semantic). I’d like to know because I’m trying to
>figure out the difference between semantic and syntactic methods, and whether
>the assumption that a sentence has a combinatorial syntax and semantics
>necessarily means that inference must at some point be syntactic, or the
>flipside, if inference was purely semantic would one have to assume that all
>logical sentences are atomic with each sentence defined independently.

        Decomposition is part of the definition of semantics; you did in
effect do that when you computed the truth table of Z.  You could even employ
syntactic transformations, like replacing  P v Q  by  ~Q -> P, and still
claim to be proceeding semantically, since you can justify them by semantics.

        The syntactic approach would be  {~Q, Z} |- ~P, or  |- (~Q & Z) -> ~P.
This doesn’t involve semantics; "|- S" means there is a sequence of formulas
a_0, a_1, …, a_n  where a_n is S, and each a_i is either a logic axiom or
arises via M.P. (i.e. there are  j, k  < i  such that  a_j  is  "a_k -> a_i").
Logic axioms are (Hilbert-Ackermann)  (A v A) -> A,  A -> (A v B),  (A v B) ->
-> (B v A),  (A -> B) ->((C v A) -> (C v B)),   for arbitrary formulas A, B, C.
Here,  A -> B  is shorthand for  ~A v B  (and  A & B  is written ~(~A v ~B) ).
There are many other ways to define  |-.

        Of course the point is that  |- S  iff  |= S.   As  |-  tends to be
cumbersome, in practice one uses |=  (or a mix);  but the two approaches
are equivalent.

                                                Ilias