Prove: if $x$ is even, then $x + 5$ is odd. I am trying to prove or disprove that if $x$ is even, then $x + 5$ is odd.
This is what I have thus far, but I am stuck:


*

*Assume that the chose variable (x) are in the domain:
(x) is an integer

*Assume the IF part of the statement:
(x) is even

*Prove the THEN part: 
Let a be a dummy variable that we assume is an integer.
$x=2a+1$ (Definition of an Odd Integer)
Now find $x+5$.
$x+5=2a+1+5$ (Because we added the same value to both sides of the equation)
$x+5=2a+6$ 
$x+5=2(a+3)$ (Because of the Distributive Property of Multiplication over Addition)
Let $z=a+3$;
$x + 5 = 2z,$ but this is the definition of an even integer; where did I mess up?

Your first line of proof does not agree with your assumption that x is even:
Assume the IF part of the statement: (x) is even
x=2a+1 (Definition of an Odd Integer)

