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Proofgold Proof

pf
Let x0 of type ι → ι be given.
Assume H0: ∀ x1 . 1eb0a.. x1 ⟶ and (SNo (x0 x1)) (∃ x2 . and (SNo x2) (∃ x3 . and (SNo x3) (∃ x4 . and (SNo x4) (∃ x5 . and (SNo x5) (∃ x6 . and (SNo x6) (∃ x7 . and (SNo x7) (∃ x8 . and (SNo x8) (x1 = bbc71.. (x0 x1) x2 x3 x4 x5 x6 x7 x8)))))))).
Assume H1: ∀ x1 . 1eb0a.. x1 ⟶ SNo (x0 x1).
Let x1 of type ι → ι be given.
Assume H2: ∀ x2 . 1eb0a.. x2 ⟶ and (SNo (x1 x2)) (∃ x3 . and (SNo x3) (∃ x4 . and (SNo x4) (∃ x5 . and (SNo x5) (∃ x6 . and (SNo x6) (∃ x7 . and (SNo x7) (∃ x8 . and (SNo x8) (x2 = bbc71.. (x0 x2) (x1 x2) x3 x4 x5 x6 x7 x8))))))).
Assume H3: ∀ x2 . 1eb0a.. x2 ⟶ SNo (x1 x2).
Let x2 of type ι be given.
Let x3 of type ι be given.
Let x4 of type ι be given.
Let x5 of type ι be given.
Let x6 of type ι be given.
Let x7 of type ι be given.
Let x8 of type ι be given.
Let x9 of type ι be given.
Assume H4: SNo x2.
Assume H5: SNo x3.
Assume H6: SNo x4.
Assume H7: SNo x5.
Assume H8: SNo x6.
Assume H9: SNo x7.
Assume H10: SNo x8.
Assume H11: SNo x9.
Apply unknownprop_959771339f7cf71f620f34cfd369621e1910f352584f3a6002a869b8d17cee3a with x0, x1, bbc71.. x2 x3 x4 x5 x6 x7 x8 x9, 50208.. x0 x1 (bbc71.. x2 x3 x4 x5 x6 x7 x8 x9) = x4 leaving 5 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H2.
Apply unknownprop_ecb826179418a88ca9938db10e4512d4b2d75d8190777ac7689e0335cc07481b with x2, x3, x4, x5, x6, x7, x8, x9 leaving 8 subgoals.
The subproof is completed by applying H4.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
The subproof is completed by applying H8.
The subproof is completed by applying H9.
The subproof is completed by applying H10.
The subproof is completed by applying H11.
Assume H12: SNo (50208.. x0 x1 (bbc71.. x2 x3 x4 x5 x6 x7 x8 x9)).
Assume H13: ∃ x10 . and (SNo x10) (∃ x11 . and (SNo x11) (∃ x12 . and (SNo x12) (∃ x13 . and (SNo x13) (∃ x14 . and (SNo x14) (bbc71.. x2 x3 x4 x5 x6 x7 x8 x9 = bbc71.. (x0 (bbc71.. x2 x3 x4 x5 x6 x7 x8 x9)) (x1 (bbc71.. x2 x3 x4 x5 x6 x7 x8 x9)) (50208.. x0 x1 (bbc71.. x2 x3 x4 x5 x6 x7 x8 x9)) x10 x11 x12 x13 x14))))).
Apply H13 with 50208.. x0 x1 (bbc71.. x2 x3 ... ... ... ... ... ...) = ....
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