Re: Answer "how to prove ~(P <--> Q) |- ~P <-->Q "



George Dance wrote:

On Jun 1, 10:37 pm, herbzet <herb...@xxxxxxxxx> wrote:
"Jesse F. Hughes" wrote:
herbzet writes:
My copy doesn't have CP. What is CP?

Conditional proof (sometimes called -> intro).

Assume P. Derive Q. Conclude P -> Q.

Thank you.


IIRC, the system Copi used in the first edition of his book on prop
logic didn't have a CP rule, and was proven to be incomplete. He used
a different rule set in the second edition.

In contrast to Jesse F. Hughes elegant proof, I offer the following
long and ***-ugly Copi-"style" proof that I've just arrived at, that
uses neither Constructive Dilemma nor Conditional Proof:

1) (P -> Q) & (R -> S) given

2) (~P v Q) & (~R v S) by definition (twice)

3) ((~P v Q) & ~R) v ((~P v Q) & S) distributive law

4) ((~P v Q) & ~R) v ((~P & S) v (Q & S)) distributive law

5) ((~P & ~R) v (Q & ~R)) v ((~P & S) v (Q & S)) distributive law

6) (~P & ~R) v (Q & ~R) v (~P & S) v (Q & S) remove parentheses

7) (~P & ~R) v [(Q & ~R) v (~P & S) v (Q & S)] restore parentheses

8) (~P & ~R) v [(~P & S) v (Q & ~R) v (Q & S)] commutative law

9) (~P & ~R) v [(~P & S) v (Q & (~R v S))] distributive law

10) (~P & ~R) v [((~P & S) v Q) & ((~P & S) v (~R v S))]
distributive law

11) (~P & ~R) v [(~P v Q) & (S v Q) & ((~P & S) v (~R v S))]
distributive law

12) (~P & ~R) v [(S v Q) & (~P v Q) & ((~P & S) v (~R v S))]
commutative law

13) (~P & ~R) v [(Q v S) & (~P v Q) & ((~P & S) v (~R v S))]
commutative law

14) (~P & ~R) v [(Q v S) & (P -> Q) & ((~P & S) v (R -> S))]
by definition (twice)

15) ~(P v R) v [(Q v S) & (P -> Q) & ((~P & S) v (R -> S))]
DeMorgan's law

16) (P v R) -> [(Q v S) & (P -> Q) & ((~P & S) v (R -> S))]
by definition

17) [(P v R)->(Q v S)] & [(P v R)->(P -> Q)] & [(P v R)->
((~P & S) v (R -> S))]
distributive law of implication into conjunction

The preceeding step is not strictly according to Copi, but it's
clearly valid. The demonstration has proceeded by equivalential
steps, and the sought-for conclusion is shown as contained in the
premise -- it is a conjunct of a formula equivalent to the premise.

If we like we can detach the conclusion with one use of the
weakening rule (P & Q) -> P

18) (P v R) -> (Q v S).

--
hz
.

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