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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιο be given.
Claim L0: ∀ x2 x3 x4 . ...(λ x5 x6 . and (x1 x5) (x5 = x6)) x2 x4x3 = x4
...
Claim L1: ∃ x2 . ∀ x3 . iff (prim1 x3 x2) (∃ x4 . and (prim1 x4 x0) (and (x1 x4) (x4 = x3)))
Apply unknownprop_aaea0f1d3f5e853f0d3287d822ec5f356e024921388ea00672dad551690ba08f with x0, λ x2 x3 . and (x1 x2) (x2 = x3).
The subproof is completed by applying L0.
Apply L1 with ∃ x2 . ∀ x3 . iff (prim1 x3 x2) (and (prim1 x3 x0) (x1 x3)).
Let x2 of type ι be given.
Assume H2: (λ x3 . ∀ x4 . iff (prim1 x4 x3) (∃ x5 . and (prim1 x5 x0) (and (x1 x5) (x5 = x4)))) x2.
Let x3 of type ο be given.
Assume H3: ∀ x4 . (∀ x5 . iff (prim1 x5 x4) (and (prim1 x5 x0) (x1 x5)))x3.
Apply H3 with x2.
Let x4 of type ι be given.
Apply H2 with x4, iff (prim1 x4 x2) (and (prim1 x4 x0) (x1 x4)).
Assume H4: prim1 x4 x2∃ x5 . and (prim1 x5 x0) (and (x1 x5) (x5 = x4)).
Assume H5: (∃ x5 . and (prim1 x5 x0) (and (x1 x5) (x5 = x4)))prim1 x4 x2.
Apply iffI with prim1 x4 x2, and (prim1 x4 x0) (x1 x4) leaving 2 subgoals.
Assume H6: prim1 x4 x2.
Apply H4 with and (prim1 x4 x0) (x1 x4) leaving 2 subgoals.
The subproof is completed by applying H6.
Let x5 of type ι be given.
Assume H7: (λ x6 . and (prim1 x6 x0) (and (x1 x6) (x6 = x4))) x5.
Apply H7 with and (prim1 x4 x0) (x1 x4).
Assume H8: prim1 x5 x0.
Assume H9: and (x1 x5) (x5 = x4).
Apply H9 with and (prim1 x4 x0) (x1 x4).
Assume H10: x1 x5.
Assume H11: x5 = x4.
Apply andI with prim1 x4 x0, x1 x4 leaving 2 subgoals.
Apply H11 with λ x6 x7 . prim1 x6 x0.
The subproof is completed by applying H8.
Apply H11 with λ x6 x7 . x1 x6.
The subproof is completed by applying H10.
Assume H6: and (prim1 x4 x0) (x1 x4).
Apply H6 with prim1 x4 x2.
Assume H7: prim1 x4 x0.
Assume H8: x1 x4.
Apply H5.
Let x5 of type ο be given.
Assume H9: ∀ x6 . and (prim1 x6 x0) (and (x1 x6) (x6 = x4))x5.
Apply H9 with x4.
Apply andI with prim1 x4 x0, and (x1 x4) (x4 = x4) leaving 2 subgoals.
The subproof is completed by applying H7.
Apply andI with x1 x4, x4 = x4 leaving 2 subgoals.
The subproof is completed by applying H8.
Let x6 of type ιιο be given.
Assume H10: x6 x4 x4.
The subproof is completed by applying H10.