Logic - Is $A \rightarrow ( B \rightarrow C) $ equivalent to $A \rightarrow C$? I know that $A  \rightarrow B$ and $B  \rightarrow C$  resolves to $A  \rightarrow C$ but does $A  \rightarrow (B  \rightarrow C)$ also resolve to $A  \rightarrow C$?
 A: No just make truthtable 
 A | B | C || A -> (B -> C) | A -> C ||
---|---|---||---M-----------|---M----||
 T | T | T ||   T     T     |   T    ||
 T | T | F ||   F     F     |   F    ||
 T | F | T ||   T     T     |   T    ||
 T | F | F ||   T     T     |   F    || DIFFERENT
 F | T | T ||   T     T     |   T    ||
 F | T | F ||   T     F     |   T    ||
 F | F | T ||   T     T     |   T    ||
 F | F | F ||   T     T     |   T    ||

For A true and B,C false, A -> (B -> C) is true while A -> C is false so they are not equivalent.
A: You can also use this argument, in order to show that the two formulas are not equivalent.
We may consider the rule :

if $\vdash A \rightarrow B$ and $\vdash B \rightarrow C$, then $\vdash A \rightarrow C$

and ask if we have also :

if $\vdash A \rightarrow (B \rightarrow C)$, then $\vdash A \rightarrow C$.

The answer is : NO.
As showed in the above truth table, the formula $(A \rightarrow (B \rightarrow C)) \rightarrow (A \rightarrow C)$ is not a tautology (for $A$ true and $B$ and $C$ false, it evaluates to false). So the above "proposed rule" is not sound. 
A: Remembering that, $$A \rightarrow  B \equiv \neg A \vee B. $$
You can reduce by DeMorgans law, $$(A \rightarrow B) \rightarrow C \equiv \neg(\neg A \vee B) \vee C$$
$$(A \wedge \neg B) \vee C.$$
Can you show that the following have the same truth values given the same inputs for A and C regardless of B? $$(A \wedge \neg B) \vee C \equiv ? A \rightarrow C.$$
A: Suppose A true, B false, and C false.  Then [A→(B→C)] is true, but (A→C) is false.
On the other hand, if [(A→B)→C] holds true, then (B→C) holds true, and {[(A→B)→C]→(B→C)} is one axiom of a three-axiom set for propositional calculus and {(A→B)→[(B→C)→(A→C)]} is another axiom in a distinct three-axiom set for propositional calculus.
A: Suppose they are equivalent. Then $\mathscr{A}=\{\neg (A\to C), A\to(B\to C)\}$ is not satisfiable. But if $\neg (A\to C)$ then we can have $A$ and $\neg C$ so if $A\to(B\to C)$, we must have $B\to C$. But $\neg C$ so $\neg B$. Thus if $A$ is true and $B$, $C$ are false, $\mathscr{A}$ is satisfied. This is a contradiction. Hence they aren't equivalent.
This argument was produced using a tableau, a systematic search for contradictions; look:
.
A: $A \to (B \to C)$ says that, if you know $A$ is true, then $B \to C$. So it says that if you know $A$ and $B$ are true, then $C$ is true.
On the other hand, $A \to C$ says that if you know $A$ is true, then $C$ must be true - with no restriction on $B$.
In general, $A \to (B\to C)$ is actually equivalent to $(A \land B) \to C$. The equivalence between these goes by the name "Currying". 
