Fill in the missing steps of the proof that (P->R)V(~Q->R)
proves P&~Q->R. Put both the missing sentence as well as
missing proof rules, sometimes you'll need to put in a missing
sentence as well as the proof rule used to derive it.
You'll need to use the following derived rule AC -- for argument
by cases:
Fill in the missing steps of the proof that (P->R)V-Q- >R) proves P&-Q->R. Put both the missing sentence as well as missing proof rules, sometimes you'll need to put in a missing sentence as well as the proof rule used to derive it. You'll need to use the following derived rule AC -- for argument by cases: Argument by Cases second form XVY ΧΑ IYA 2 AC In words, if in a derivation you already have a sentence of the form XvY. subderivation from X as assumption to Z as conclusion, and a second sub derivation from Y as assumption to Z as conclusion, you are licensed to write Z as a condusion anywhere below. (P- >R)V-Q- >R) Assumption P&-Q Assumption Assumption P->R ~Q->R Assumption (P&-0)- >R
Argument by Cases (second form) XVY F AC In words, if in a derivation you already have a sentence of the form XvY, a subderivation from X as assumption to Z as conclusion, and a second sub- derivation from Y as assumption to Z as conclusion, you are licensed to write Z as a condusion anywhere below.
Fill in the missing steps of the proof that (P->R)V-Q- >R) proves P&-Q->R. Put both the missing sentence as well as miss
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Fill in the missing steps of the proof that (P->R)V-Q- >R) proves P&-Q->R. Put both the missing sentence as well as miss
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