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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιιι be given.
Let x2 of type ιιι be given.
Let x3 of type ιιι be given.
Let x4 of type ι be given.
Let x5 of type ιιι be given.
Let x6 of type ιιιι be given.
Let x7 of type ιιι be given.
Let x8 of type ιιιι be given.
Let x9 of type ιιιι be given.
Let x10 of type ιιι be given.
Let x11 of type ιιι be given.
Let x12 of type ιιι be given.
Let x13 of type ιιι be given.
Let x14 of type ι be given.
Assume H0: In x14 x0.
Let x15 of type ι be given.
Assume H1: In x15 x0.
Let x16 of type ι be given.
Assume H2: In x16 x0.
Let x17 of type ι be given.
Assume H3: In x17 x0.
Let x18 of type ι be given.
Assume H4: In x18 x0.
Assume H5: Loop_with_defs x0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13.
Assume H6: not (x6 x18 (x1 (x3 x4 x14) (x9 x15 x16 x14)) x17 = x4).
Apply unknownprop_f7b5ec75bad85de491ae811af7281b8523890ceb7b0a21e077fce8976ff58a01 with x0, x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13 leaving 2 subgoals.
The subproof is completed by applying H5.
Let x19 of type ο be given.
Assume H7: ∀ x20 . and (In x20 x0) (∃ x21 . and (In x21 x0) (∃ x22 . and (In x22 x0) (∃ x23 . and (In x23 x0) (∃ x24 . and (In x24 x0) (not (x6 x24 (x1 (x3 x4 x20) (x9 x21 x22 x20)) x23 = x4))))))x19.
Apply H7 with x14.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with In x14 x0, ∃ x20 . and (In x20 x0) (∃ x21 . and (In x21 x0) (∃ x22 . and (In x22 x0) (∃ x23 . and (In x23 x0) (not (x6 x23 (x1 (x3 x4 x14) (x9 x20 x21 x14)) x22 = x4))))) leaving 2 subgoals.
The subproof is completed by applying H0.
Let x20 of type ο be given.
Assume H8: ∀ x21 . and (In x21 x0) (∃ x22 . and (In x22 x0) (∃ x23 . and (In x23 x0) (∃ x24 . and (In x24 x0) (not (x6 x24 (x1 (x3 x4 x14) (x9 x21 x22 x14)) x23 = x4)))))x20.
Apply H8 with x15.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with In x15 x0, ∃ x21 . and (In x21 x0) (∃ x22 . and (In x22 x0) (∃ x23 . and (In x23 x0) (not (x6 x23 (x1 (x3 x4 x14) (x9 x15 x21 x14)) x22 = x4)))) leaving 2 subgoals.
The subproof is completed by applying H1.
Let x21 of type ο be given.
Assume H9: ∀ x22 . and (In x22 x0) (∃ x23 . and (In x23 x0) (∃ x24 . and (In x24 x0) (not (x6 x24 (x1 (x3 x4 x14) (x9 x15 x22 x14)) x23 = x4))))x21.
Apply H9 with x16.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with In x16 x0, ∃ x22 . and (In x22 x0) (∃ x23 . ...) leaving 2 subgoals.
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