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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιιι be given.
Assume H0: ∀ x2 . x2x0∀ x3 . x3x0x1 x2 x3x0.
Assume H1: explicit_Group x0 x1.
Let x2 of type ι be given.
Let x3 of type ιιι be given.
Assume H2: ∀ x4 . x4x2∀ x5 . x5x2x3 x4 x5x2.
Assume H3: explicit_Group x2 x3.
Let x4 of type ιιι be given.
Assume H4: ∀ x5 . x5setprod x0 x2∀ x6 . x6setprod x0 x2lam 2 (λ x7 . If_i (x7 = 0) (x1 (ap x5 0) (ap x6 0)) (x3 (ap x5 1) (ap x6 1))) = x4 x5 x6.
Apply H1 with explicit_Group (setprod x0 x2) x4.
Assume H5: and (∀ x5 . x5x0∀ x6 . x6x0x1 x5 x6x0) (∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x1 x5 (x1 x6 x7) = x1 (x1 x5 x6) x7).
Apply H5 with (∃ x5 . and (x5x0) (and (∀ x6 . x6x0and (x1 x5 x6 = x6) (x1 x6 x5 = x6)) (∀ x6 . x6x0∃ x7 . and (x7x0) (and (x1 x6 x7 = x5) (x1 x7 x6 = x5)))))explicit_Group (setprod x0 x2) x4.
Assume H6: ∀ x5 . x5x0∀ x6 . x6x0x1 x5 x6x0.
Assume H7: ∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x1 x5 (x1 x6 x7) = x1 (x1 x5 x6) x7.
Assume H8: ∃ x5 . and (x5x0) (and (∀ x6 . x6x0and (x1 x5 x6 = x6) (x1 x6 x5 = x6)) (∀ x6 . x6x0∃ x7 . and (x7x0) (and (x1 x6 x7 = x5) (x1 x7 x6 = x5)))).
Apply H3 with explicit_Group (setprod x0 x2) x4.
Assume H9: and (∀ x5 . x5x2∀ x6 . x6x2x3 x5 x6x2) (∀ x5 . ...∀ x6 . ...∀ x7 . ...x3 x5 (x3 x6 x7) = x3 (x3 x5 x6) ...).
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