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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?

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

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