Formal logic to confirm $x+1 = 1 +x$

So i've been stuck on this trouble for concerning a hr. I can not identify just how to do it, and also aid would certainly be definitely impressive. This is what is offered :

$0 + 1 = 1$

$\forall x (x + 0 = x)$

$\forall x \forall y [x + (y+1)= (x+y)+1]$

$[0 + 1 = 1 + 0 \wedge \forall x(x + 1 = 1 + x \rightarrow (x+1)+1 = 1+(x+1)))] \rightarrow \forall x (x+1=1+x)$

And with these properties, I require to get $\forall x (x+1=1+x)$.

Aid would certainly be wonderful. I maintain obtaining closeish, just to locate I'm in fact doing it entirely incorrect.

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2019-12-02 02:49:34
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Answers: 2

The last axiom is primarily your induction declaration : it claims that if you can confirm that $x+1 = 1+x$ for $x=0$, and also you can confirm that if it holds for $x$ after that it holds for $x+1$, after that you can end that the declaration $x+1=1+x$ holds for all $x$. So you intend to utilize your first and also 2nd axioms, along with your 3rd axiom, to show that the properties of the last axiom hold. That will certainly offer the verdict you desire.

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2019-12-03 01:13:21
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Step 1 : Prove 0+1 = 1+0

Apply property # 2 for x = 1, and also replace right into property # 1.

Action 2 : Prove Ax [x+1 = 1+x - > x+1+1 = 1+(x+1) ]

Suppose k+1 = 1+k (property)

Apply property # 3 for x = 1, y = k. Substitute from last property. Apply proportion of = to get k+1+1 = 1+(k+1).

Generalise. Apply property # 4 to get called for outcome.

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2019-12-03 00:54:50
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