Identities in first order logic

Question

When given the statement (∀) A() → B() → C, can I rewrite it as (∀) (A() ^ B()) → C ? Also, if so, how can I can prove it please? Thanks!

Answer 1

Yes, indeed, $(\forall x)~(A(x)\to(B(x)\to C)) \iff (\forall x)~((A(x)\wedge B(x))\to C)$

We may derive $C$ when we assume that $(\forall x)~(A(x)\to(B(x)\to C))$ and, for an arbitrary entity $c$, that $A(c)\wedge B(c)$.

Therefore $(\forall x)~(A(x)\to(B(x)\to C)) \implies (\forall x)~((A(x)\wedge B(x))\to C)$

We may derive $C$ when we assume that $(\forall x)~(A(x)\wedge B(x))\to C$, and, for an arbitrary entity $c$, that $A(c)$ and $B(c)$.

Therefore $(\forall x)~(A(x)\to(B(x)\to C)) \impliedby (\forall x)~((A(x)\wedge B(x))\to C)$

Answer 2

OK so let's start by asking what it would take to make $(\forall x) A(x) \rightarrow B(x) \rightarrow C $ false. Well we would need $\forall x B(x) \rightarrow C$ to be false and $(\forall x) A(x)$ to be true. In turn $\forall x B(x) \rightarrow C$ is false if and only if $\forall x B(x)$ is true and $C$ is false. Now what would it take for $(\forall x) (A(x) \wedge B(x)) \rightarrow C$ to be false? Again we would need $C$ to be false and $(\forall x) (A(x) \wedge B(x))$ to be true. This only happens if, for every $x$ $A$ is true and $B$ is true. So the conditions for both statements being false are identical hence the conditions for both statements being true are identical and the statements are equivalent.

If you have learned the tableau method then you will recognise this as an informal application of it. If not, then there is no need to worry, the reasoning still stands alone.