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

[Vibhukanishk]

What to say about this?

Is the logic used in the solution supports the statement?

 

[TeethWhitener]

No. You just asserted that $g(y)=h(y)^n$ with no justification.

 

[fresh_42]

There is quite a bit of information missing. E.g., there is no quantifier on $h(y)$ in the initial statement.

 

[Vibhukanishk]

No. You just asserted that $g(y)=h(y)^n$ with no justification.

 

[TeethWhitener]

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.