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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ιι be given.
Assume H0: Subq x0 x1.
Assume H1: ∀ x3 . In x3 x1nIn x3 x0In 0 (x2 x3).
Apply unknownprop_c3fe42b21df0810041479a97b374de73f7754e07c8af1c88386a1e7dc0aad10f with Pi x0 (λ x3 . x2 x3), Pi x1 (λ x3 . x2 x3).
Let x3 of type ι be given.
Assume H2: In x3 (Pi x0 (λ x4 . x2 x4)).
Apply unknownprop_c20579f7ec03c9b411c1afcdcbd6eb7f887b4dea35b13dd2fe5a71172b6554fe with x0, x2, x3, In x3 (Pi x1 (λ x4 . x2 x4)) leaving 2 subgoals.
The subproof is completed by applying H2.
Assume H3: ∀ x4 . In x4 x3and (setsum_p x4) (In (ap x4 0) x0).
Assume H4: ∀ x4 . In x4 x0In (ap x3 x4) (x2 x4).
Apply unknownprop_1a1eed9c2e0652a509eabe7b8f07e31768cab0357ad1d97cb464202e3d371a17 with x1, x2, x3 leaving 2 subgoals.
Let x4 of type ι be given.
Assume H5: In x4 x3.
Apply andE with setsum_p x4, In (ap x4 0) x0, and (setsum_p x4) (In (ap x4 0) x1) leaving 2 subgoals.
Apply H3 with x4.
The subproof is completed by applying H5.
Assume H6: setsum_p x4.
Assume H7: In (ap x4 0) x0.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with setsum_p x4, In (ap x4 0) x1 leaving 2 subgoals.
The subproof is completed by applying H6.
Apply unknownprop_cc8f63ddfbec05087d89028647ba2c7b89da93a15671b61ba228d6841bbab5e9 with x0, x1, ap x4 0 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H7.
Let x4 of type ι be given.
Assume H5: In x4 x1.
Apply unknownprop_80056f9db85f84f8ce0644e1cdb62f5e66ec62c041f28dd7a07b3c46de1ea696 with x4, x0, In (ap x3 x4) (x2 x4) leaving 2 subgoals.
Assume H6: In x4 x0.
Apply H4 with x4.
The subproof is completed by applying H6.
Assume H6: nIn x4 x0.
Claim L7: ap x3 x4 = 0
Apply unknownprop_1d401bfd2bdb443cb4ed66a0412f4299139e285cca5e460f0f1c03467ae8078c with x0, x2, x3, λ x5 x6 . ap x5 x4 = 0 leaving 2 subgoals.
The subproof is completed by applying H2.
Apply unknownprop_65bac37c150594021865ca4bcc52ede90a62ee904e2b827d4b233886c39d597e with x0, ap x3, x4.
The subproof is completed by applying H6.
Apply L7 with λ x5 x6 . In x6 (x2 x4).
Apply H1 with x4 leaving 2 subgoals.
The subproof is completed by applying H5.
The subproof is completed by applying H6.