Thread: Prop Logic shenanigans

A, B, and C are arbitrary wffs, how would I prove that:

{A->B, B->C} |- (A->C)

Originally Posted by snowtone

A, B, and C are arbitrary wffs, how would I prove that:

{A->B, B->C} |- (A->C)

1) Α===>Β............................................ ......................given

2) B====>C........................................... ........................given

3) A................................................. ...........................assumption

4) B................................................. ..........................1,3 M.Ponens

5) C................................................. ...........................2,4 M.Ponens

6) A=====>C.......................................... ..................BY using the deduction theorem or called differently the law of conditional proof
from the lines 3 to 5

Hi -

Originally Posted by snowtone

A, B, and C are arbitrary wffs, how would I prove that:

{A->B, B->C} |- (A->C)

As an alternative, set up a truth table for the proposition



and show that this is a tautology.

Grandad

Why not for the proposition:

[(A---->B)&(B---->C)]====>(A---->C)

Hello archidi -

Originally Posted by archidi

Why not for the proposition:

[(A---->B)&(B---->C)]====>(A---->C)

No particular reason. This works just as well.

Grandad