Logic proof for $(x\leq y\to 0\leq y-x), x+y-y=x$, and the property that the product two consecutive numbers is even

Question

I need to find the proof for

• $(x\leq y\to 0\leq y-x)$
• $x+y-y=x$
• $(\exists u ((u+u)=s(x)\times x))$, where s is the successor function. Or, the product of two consecutive natural numbers is even.

I have no idea how where to start for the first two, and I can see the last statement needing to use Peano logic, but how to actually do it is completely beyond me.

None of these statements have universal quantifiers, so I don't think I can employ induction.

Answer 1

Hint

If we work with Peano arithmetic, we must define $\le$:

$x \le y \text { iff } \exists z \ (y=z+x)$.

If so, $y-x$ is the unique $z$ such that $y=z+x$.

Clearly, $z \ge 0$ (every natural number is so).

For the second one, $x+y-y=x+(y-y)$. But $y-y=0$, because $y=y+0$.

Thus: $x+y-y=x+(y-y)=x+0=x$.

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For the third one, we can use induction, as well as the axioms for $\times$.

i) Basis: $0 \times s(0)=0 \times 0 + 0 = 0+0$.

ii) Induction: assume $s(x) \times x=x \times x + x= u + u$, for some $u$.

Now: $s(x+1) \times (x+1) = (x+1) \times (x+1) + (x+1) = (x+1) \times x + (x+1) + (x+1) = (x \times x + x) + (x+1) + (x+1)$.

But we know that $x \times x + x= u + u$, and thus:

$s(x+1) \times (x+1) = (x \times x + x) + (x+1) + (x+1) = u + u + (x+1) + (x+1) = (u + (x+1)) + (u + (x+1)).$