One Response to “question about semantic methods”

1. admin says:

   January 26 at 2:35 am

   In article <6av5ti$3c…@morgoth.sfu.ca>,

   - Hide quoted text — Show quoted text -

   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