# Modus ponens rule

Prove for the calculus of propositions: $$A\rightarrow B\mapsto(B\rightarrow C)\rightarrow(A\rightarrow C)$$ I had used axioms and most suitable was this one: $$X\rightarrow Y\rightarrow(X\rightarrow(Y\rightarrow Z)\rightarrow(X\rightarrow Z))$$ (axiom2). Then I used Modus ponens rule and got this result: $$A\rightarrow B,\ A\rightarrow B\rightarrow (A\rightarrow(B\rightarrow C)\rightarrow(A\rightarrow C))\\ A\rightarrow(B\rightarrow C)\rightarrow(A\rightarrow C)$$

Then I didn't understand, how to get $$(B\rightarrow C)\rightarrow(A\rightarrow C)$$. What axiom should I use or rule?

Using the Deduction Theorem:

1. $$A \to B$$ --- premise

2. $$B \to C$$ --- assumed [a]

3. $$(B \to C) \to (A \to (B \to C))$$ --- Ax.1

4. $$A \to (B \to C)$$ --- from 2) and 3) by Modus Ponens

5. $$(A \to B) \to ((A \to (B \to C)) \to (A \to C))$$ --- Ax.2

6. $$((A \to (B \to C)) \to (A \to C))$$ --- from 1) and 5) by Modus Ponens

7. $$A \to C$$ --- from 4) and 6) by Modus Ponens.

1. $$(B \to C) \to (A \to C)$$ --- from 2) and 7) by DT, discharging assumption [a]
• i have a question: could we use Deduction Theorem for this one:$A→B,B→C↦A→C$? Jun 17 at 18:44
• @sjr_25 - YES. Modify the above proof assuming also $A$, followed by MP twice to get $C$ and then conclude by DT with $A \to C$. Jun 18 at 5:50
• Can I ask another question on this topic? Jun 18 at 6:26
• @sjr_25 - why not ? Jun 18 at 6:43
• Can i rewrite this ¬ A,¬ B→(¬ AVB)→(AV¬B) like A⟷B Jun 18 at 6:58