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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιιιο be given.
Assume H0: ∀ x2 . x1 c4def.. x2 x2.
Assume H1: ∀ x2 x3 x4 x5 x6 . x1 x2 x4 x5x1 x3 x5 x6x1 (6b90c.. x2 x3) x4 x6.
Assume H2: ∀ x2 . x1 c9248.. x2 236c6...
Assume H3: ∀ x2 x3 x4 . x1 x2 x3 x4x1 (a6e19.. x2) x3 (0b8ef.. x4).
Assume H4: ∀ x2 x3 x4 . x1 x2 x3 x4x1 (2fe34.. x2) x3 (6c5f4.. x4).
Assume H5: ∀ x2 x3 x4 x5 x6 . e05e6.. x3x1 x2 (cfc98.. x4 x5) x6x1 (3e00e.. x2 x3) (cfc98.. (0b8ef.. x4) x5) x6.
Assume H6: ∀ x2 x3 x4 x5 x6 . e05e6.. x2x1 x3 (cfc98.. x4 x5) x6x1 (3e00e.. x2 x3) (cfc98.. (6c5f4.. x4) x5) x6.
Assume H7: ∀ x2 x3 x4 x5 x6 . x1 x2 x4 x5x1 x3 x4 x6x1 (f9341.. x2 x3) x4 (cfc98.. x5 x6).
Assume H8: ∀ x2 x3 x4 x5 . x1 x2 x3 x5x1 (1fa6d.. x2) (cfc98.. x3 x4) x5.
Assume H9: ∀ x2 x3 x4 x5 . x1 x2 x4 x5x1 (3a365.. x2) (cfc98.. x3 x4) x5.
The subproof is completed by applying H0 with x0.