I Does Defining ##g(y)## as ##h(y)^n## Validate the Statement?

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The discussion centers on the validity of defining g(y) as h(y)^n in relation to a specific statement. Participants argue that the logic supporting this definition lacks justification and is insufficiently detailed. The absence of a quantifier on h(y) is highlighted as a critical omission. It is emphasized that while defining g(y) in this manner makes the statement trivially true, it does not constitute a proof. Overall, the consensus is that the assertion does not validate the statement due to its reliance on definitions rather than logical reasoning.
Vibhukanishk
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What to say about this?
1659958487965.png

Is the logic used in the solution supports the statement?
 
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No. You just asserted that ##g(y)=h(y)^n## with no justification.
 
There is quite a bit of information missing. E.g., there is no quantifier on ##h(y)## in the initial statement.
 
TeethWhitener said:
No. You just asserted that ##g(y)=h(y)^n## with no justification.
IMG_20220808_182840.jpg
 
If you define ##g(y)## as ##h(y)^n##, then of course it's true, but there's also nothing to prove; it's all definitions.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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