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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)

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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 $\mathsf{False}$

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    I'm sorry, I must be misunderstanding something. Why can you just set the propositions to be arbitrary truth values? What if I were try to prove a statement $A\Rightarrow B$ and right away I just say that $A$ is true and $B$ is false, thus the statement is incorrect and we cannot proceed? – 2017-02-16
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    @jonem: Consider the following analogous siutation: it's as if you said you were trying to prove $$\large a+b=ac +d$$ and I said, *"That formula is false, since if $a=2$ and $b=1$ and $c=5$ and $d=9$, the left side is $3$ and the right side is $19$"*. – 2017-02-16
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    You're claiming that a formula / statement is valid ***in general***, and I'm pointing out that since there's at least one ***specific*** case where it's not, then your claim is incorrect. – 2017-02-16