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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι → ι be given.
Assume H0: ∀ x2 . prim1 x2 (e5b72.. x0) ⟶ prim1 (x1 x2) (e5b72.. x0).
Assume H1: ∀ x2 . prim1 x2 (e5b72.. x0) ⟶ ∀ x3 . prim1 x3 (e5b72.. x0) ⟶ Subq x2 x3 ⟶ Subq (x1 x2) (x1 x3).
Claim L2: ...
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Claim L3: ...
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Claim L4: ...
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Claim L5: Subq (x1 (1216a.. x0 (λ x2 . ∀ x3 . prim1 x3 (e5b72.. x0) ⟶ Subq (x1 x3) x3 ⟶ prim1 x2 x3))) (1216a.. x0 (λ x2 . ∀ x3 . prim1 ... ... ⟶ Subq (x1 x3) x3 ⟶ prim1 x2 x3))
...
Claim L6: Subq (x1 (x1 (1216a.. x0 (λ x2 . ∀ x3 . prim1 x3 (e5b72.. x0) ⟶ Subq (x1 x3) x3 ⟶ prim1 x2 x3)))) (x1 (1216a.. x0 (λ x2 . ∀ x3 . prim1 x3 (e5b72.. x0) ⟶ Subq (x1 x3) x3 ⟶ prim1 x2 x3)))
Apply H1 with x1 (1216a.. x0 (λ x2 . ∀ x3 . prim1 x3 (e5b72.. x0) ⟶ Subq (x1 x3) x3 ⟶ prim1 x2 x3)), 1216a.. x0 (λ x2 . ∀ x3 . prim1 x3 (e5b72.. x0) ⟶ Subq (x1 x3) x3 ⟶ prim1 x2 x3) leaving 3 subgoals.
The subproof is completed by applying L3.
The subproof is completed by applying L2.
The subproof is completed by applying L5.
Let x2 of type ο be given.
Assume H7: ∀ x3 . and (prim1 x3 (e5b72.. x0)) (x1 x3 = x3) ⟶ x2.
Apply H7 with 1216a.. x0 (λ x3 . ∀ x4 . prim1 x4 (e5b72.. x0) ⟶ Subq (x1 x4) x4 ⟶ prim1 x3 x4).
Apply andI with prim1 (1216a.. x0 (λ x3 . ∀ x4 . prim1 x4 (e5b72.. x0) ⟶ Subq (x1 x4) x4 ⟶ prim1 x3 x4)) (e5b72.. x0), x1 (1216a.. x0 (λ x3 . ∀ x4 . prim1 x4 (e5b72.. x0) ⟶ Subq (x1 x4) x4 ⟶ prim1 x3 x4)) = 1216a.. x0 (λ x3 . ∀ x4 . prim1 x4 (e5b72.. x0) ⟶ Subq (x1 x4) x4 ⟶ prim1 x3 x4) leaving 2 subgoals.
The subproof is completed by applying L2.
Apply set_ext with x1 (1216a.. x0 (λ x3 . ∀ x4 . prim1 x4 (e5b72.. x0) ⟶ Subq (x1 x4) x4 ⟶ prim1 x3 x4)), 1216a.. x0 (λ x3 . ∀ x4 . prim1 x4 (e5b72.. x0) ⟶ Subq (x1 x4) x4 ⟶ prim1 x3 x4) leaving 2 subgoals.
The subproof is completed by applying L5.
Apply L4 with x1 (1216a.. x0 (λ x3 . ∀ x4 . prim1 x4 (e5b72.. x0) ⟶ Subq (x1 x4) x4 ⟶ prim1 x3 x4)) leaving 2 subgoals.
The subproof is completed by applying L3.
The subproof is completed by applying L6.
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