Prove $ A \supset B , B \supset C \vdash A \supset C $

Question

From Kleene's Introduction to Metamathematics, page 94 :

"as further examples [of deduction] the reader may establish : "

$ A \supset B , B \supset C \vdash A \supset C $.

(Here $\supset$ means logical implication, not superset).

Answer 1

Actually you don't really need the deduction theorem: it just makes the proof easier.

Here is an alternative to Makholm's hint:

1. $A⊃B$, premise

2. $B⊃C$, premise

3. $(A⊃B)⊃((A⊃(B⊃C))⊃(A⊃C))$, Axiom Schema 1b (see p.82)

4. $((A⊃(B⊃C))⊃(A⊃C))$, 1,3, Modus Ponens

5. $(B⊃C) ⊃ (A⊃(B⊃C))$, Axiom Schema 1a

6. $(A⊃(B⊃C))$, 2,5, Modus Ponens

7. $(A⊃C$, 4,6, Modus Ponens

Next time make explicit your proof system in the body of the question:

This way you can get people who doesn't have access to the book you refer to help you too.

Answer 2

Since you have the Deduction Theorem, you just need to prove $A\supset B, B\supset C, A\vdash C$.

Use modus ponens on $A$ and $A\supset B$ to get $B$. Then modus ponens again with $B\supset C$ to get $C$.