Search for blocks/addresses/...

Proofgold Proof

pf
Let x0 of type ι(ι((ιο) → ο) → ο) → ((ιο) → ο) → ο be given.
Let x1 of type ι be given.
Let x2 of type ((ιο) → ο) → ο be given.
Assume H0: b9fc2.. x0 x1 x2.
Apply H0 with λ x3 . λ x4 : ((ι → ο) → ο) → ο . ∃ x5 : ι → ((ι → ο) → ο) → ο . and (∀ x6 . x6x3b9fc2.. x0 x6 (x5 x6)) (x4 = x0 x3 x5).
Let x3 of type ι be given.
Let x4 of type ι((ιο) → ο) → ο be given.
Assume H1: ∀ x5 . x5x3∃ x6 : ι → ((ι → ο) → ο) → ο . and (∀ x7 . x7x5b9fc2.. x0 x7 (x6 x7)) (x4 x5 = x0 x5 x6).
Let x5 of type ο be given.
Assume H2: ∀ x6 : ι → ((ι → ο) → ο) → ο . and (∀ x7 . x7x3b9fc2.. x0 x7 (x6 x7)) (x0 x3 x4 = x0 x3 x6)x5.
Apply H2 with x4.
Apply andI with ∀ x6 . x6x3b9fc2.. x0 x6 (x4 x6), x0 x3 x4 = x0 x3 x4 leaving 2 subgoals.
Let x6 of type ι be given.
Assume H3: x6x3.
Apply H1 with x6, b9fc2.. x0 x6 (x4 x6) leaving 2 subgoals.
The subproof is completed by applying H3.
Let x7 of type ι((ιο) → ο) → ο be given.
Assume H4: and (∀ x8 . x8x6b9fc2.. x0 x8 (x7 x8)) (x4 x6 = x0 x6 x7).
Apply H4 with b9fc2.. x0 x6 (x4 x6).
Assume H5: ∀ x8 . x8x6b9fc2.. x0 x8 (x7 x8).
Assume H6: x4 x6 = x0 x6 x7.
Apply H6 with λ x8 x9 : ((ι → ο) → ο) → ο . b9fc2.. x0 x6 x9.
Apply unknownprop_19e14cfefa9c62802c3472f7b613074ec9e1720289fb25abc9d15e165cea68de with x0, x6, x7.
The subproof is completed by applying H5.
Let x6 of type (((ιο) → ο) → ο) → (((ιο) → ο) → ο) → ο be given.
Assume H3: x6 (x0 x3 x4) (x0 x3 x4).
The subproof is completed by applying H3.