Prove this statement Let's say we have some predicate $ P $, then how does one prove the following statement?
$$\forall x \forall y ((x=y \wedge P (x))\Rightarrow P (y)) $$
 A: Direct Proof
As $x=y$, $\forall x ((x=x \wedge P (x))\Rightarrow P (x))$ by substitution, and $\forall x ((x=x \wedge P (x))\Rightarrow P (x))$ because by reflexitivity of equality, $\forall x (P (x)\Rightarrow P (x))$ by conjunction elimination, and the result is true.

Proof by Contradiction
Let's try to prove by contradiction, and suppose that's false.
If so, $\exists x \exists y ((x=y \wedge P (x))\not\Rightarrow P (y))$ (read out loud and you'll see it is obvious). 
So $\exists x ((x=x \wedge P (x))\not\Rightarrow P (x))$, and $\exists x (P (x)\not\Rightarrow P (x))$  which is obviously a contradiction.

Proof through a Truth Table
$x=y$ is a statement, so $(0; 1)$ and $(1; 0)$ values, respectivly, for $x$ and $y$, don't apply. 
P(x) P(y)  → (see http://en.wikipedia.org/wiki/Material_conditional)
0    0     1
1    1     1

(where $1$ is true and $0$ is false)
Therefore $\forall x \forall y ((x=y \wedge P (x))\Rightarrow P (y))$ follows $\forall x \forall y ((x=y \wedge 1)\Rightarrow 1)$ and $1 \Rightarrow 1$, $1$, and $1$ is true-
A: If you read up about First Order Logic with Equality, your statement follows readily using the Substitution for formulas axiom scheme, aka. Leibniz' Law
