1. ## Prop Logic shenanigans

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

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

2. 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

3. ## Logic proofs

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

$A \wedge((A \implies B) \wedge (B \implies C)) \implies C$

and show that this is a tautology.

4. Why not for the proposition:

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

5. ## Logic proofs

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.