Step
*
of Lemma
not-poly-choice-eta-2
¬(∀f:Base. ((∀x,y:Base. ((f x y) = x ∈ Base))
⇒ (f ~ λx,y. (f x y))))
BY
{ Assert ⌜∀x,y:Base. (eval z = x in λy.z y ~ x)⌝⋅ }
1
.....assertion.....
∀x,y:Base. (eval z = x in λy.z y ~ x)
2
1. ∀x,y:Base. (eval z = x in λy.z y ~ x)
⊢ ¬(∀f:Base. ((∀x,y:Base. ((f x y) = x ∈ Base))
⇒ (f ~ λx,y. (f x y))))
Latex:
Latex:
\mneg{}(\mforall{}f:Base. ((\mforall{}x,y:Base. ((f x y) = x)) {}\mRightarrow{} (f \msim{} \mlambda{}x,y. (f x y))))
By
Latex:
Assert \mkleeneopen{}\mforall{}x,y:Base. (eval z = x in \mlambda{}y.z y \msim{} x)\mkleeneclose{}\mcdot{}
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