How would I go about proving $(A \Rightarrow B) \Rightarrow ((A \wedge C)\Rightarrow D)$?

Question

Does this expression reduce down to something simpler? How would I go about showing this?

Wolfram alpha gives: http://www.wolframalpha.com/input/?i=(A+implies+B)+implies+((C+AND+A)+implies+D)

Answer 1

What you've written is incorrect. For example, suppose $A=B=C=\mathsf{True}$ and $D=\mathsf{False}$. Then:

• The statement $A\implies B$ is $\mathsf{True}\implies \mathsf{True}$, which is $\mathsf{True}$

• The statement $A\wedge C$ is $\mathsf{True}\wedge \mathsf{True}$, which is $\mathsf{True}$

• Therefore the statement $(A\wedge C)\implies D$ is $\mathsf{True}\implies \mathsf{False}$, which is $\mathsf{False}$

• Therefore the statement $(A\implies B)\implies ((A\wedge C)\implies D)$ is $\mathsf{True}\implies \mathsf{False}$, which is $\mathsf{False}$