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Proofgold Proof

pf
Let x0 of type ιο be given.
Let x1 of type ιιι be given.
Let x2 of type ιιι be given.
Assume H0: ∀ x3 x4 . x0 x3x0 x4x0 (x1 x3 x4).
Assume H1: ∀ x3 x4 x5 . x0 x3x0 x4x0 x5x2 x3 (x1 x4 x5) = x1 (x2 x3 x4) (x2 x3 x5).
Assume H2: ∀ x3 x4 x5 . x0 x3x0 x4x0 x5x2 (x1 x3 x4) x5 = x1 (x2 x3 x5) (x2 x4 x5).
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.
Let x10 of type ι be given.
Let x11 of type ι be given.
Assume H3: x0 x3.
Assume H4: x0 x4.
Assume H5: x0 x5.
Assume H6: x0 x6.
Assume H7: x0 x7.
Assume H8: x0 x8.
Assume H9: x0 x9.
Assume H10: x0 x10.
Assume H11: x0 x11.
Apply unknownprop_f8cd2cc9fa3527ece23bf769ef6e4562a4db8193110bd03a6fde0a07ab4c38b8 with x0, x1, x2, x3, x4, x5, x6, x1 x7 (x1 x8 (x1 x9 (x1 x10 x11))), λ x12 x13 . x13 = x1 (x1 (x2 x3 x7) (x1 (x2 x3 x8) (x1 (x2 x3 x9) (x1 (x2 x3 x10) (x2 x3 x11))))) (x1 (x1 (x2 x4 x7) (x1 (x2 x4 x8) (x1 (x2 x4 x9) (x1 (x2 x4 x10) (x2 x4 x11))))) (x1 (x1 (x2 x5 x7) (x1 (x2 x5 x8) (x1 (x2 x5 x9) (x1 (x2 x5 x10) (x2 x5 x11))))) (x1 (x2 x6 x7) (x1 (x2 x6 x8) (x1 (x2 x6 x9) (x1 (x2 x6 x10) (x2 x6 x11))))))) leaving 8 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
Apply unknownprop_d7ce6357a8261c6a4be44f579bedcb1c2d65cec14964ea078af8f02cc5aab85a with x0, x1, x7, x8, x9, x10, x11 leaving 6 subgoals.
The subproof is completed by applying H0.
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.
Apply unknownprop_ba42d8afab82952b0d7dfc0b0a5aca8e776863449ac4fa660349884f276c83d1 with x0, x1, x2, x7, x8, x9, x10, x11, x3, λ x12 x13 . x1 x13 (x1 (x2 x4 (x1 x7 (x1 x8 (x1 x9 (x1 x10 x11))))) (x1 (x2 x5 (x1 x7 (x1 x8 (x1 x9 (x1 x10 x11))))) (x2 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x10 x11))))))) = x1 (x1 (x2 x3 x7) (x1 (x2 x3 x8) (x1 (x2 x3 x9) (x1 (x2 x3 x10) (x2 x3 x11))))) (x1 (x1 (x2 x4 x7) (x1 ... ...)) ...) leaving 9 subgoals.
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