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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 ι be given.
Assume H0: In (setsum 1 x2) (setsum x0 x1).
Apply unknownprop_eb8e8f72a91f1b934993d4cb19c84c8270f73a3626f3022b683d960a7fef89cb with ∃ x3 . and (In x3 x0) (setsum 1 x2 = setsum 0 x3), ∃ x3 . and (In x3 x1) (setsum 1 x2 = setsum 1 x3), In x2 x1 leaving 3 subgoals.
Apply unknownprop_c529973cb32f8d02a3950eda53e547d40c4a0e8faca1777353233a3377534f09 with x0, x1, setsum 1 x2.
The subproof is completed by applying H0.
Apply unknownprop_74210cc9b2960bdcb3eab56d0b4f5ba5a5771478f68cd794d919dafbcd157b00 with λ x3 x4 : ι → ι . (∃ x5 . and (In x5 x0) (setsum 1 x2 = x3 x5))In x2 x1.
Apply unknownprop_e88b17fc7534a834a3292f38867e04234c9b0d119c42f884c32fbabae05b0d7e with λ x3 x4 : ι → ι . (∃ x5 . and (In x5 x0) (x3 x2 = Inj0 x5))In x2 x1.
Assume H1: ∃ x3 . and (In x3 x0) (Inj1 x2 = Inj0 x3).
Apply FalseE with In x2 x1.
Apply unknownprop_3848cfb1fd522cb609408da39f227a9c05924a24919f21041d0880590b824ef5 with λ x3 . In x3 x0, λ x3 . Inj1 x2 = Inj0 x3, False leaving 2 subgoals.
The subproof is completed by applying H1.
Let x3 of type ι be given.
Assume H2: In x3 x0.
Assume H3: Inj1 x2 = Inj0 x3.
Apply notE with Inj0 x3 = Inj1 x2 leaving 2 subgoals.
The subproof is completed by applying unknownprop_2909aa42e9d0a354d060bc7d707070a586f9ab4666ef2c2e92d5cb1072a37e98 with x3, x2.
Let x4 of type ιιο be given.
The subproof is completed by applying H3 with λ x5 x6 . x4 x6 x5.
Apply unknownprop_e88b17fc7534a834a3292f38867e04234c9b0d119c42f884c32fbabae05b0d7e with λ x3 x4 : ι → ι . (∃ x5 . and (In x5 x1) (x3 x2 = x3 x5))In x2 x1.
Assume H1: ∃ x3 . and (In x3 x1) (Inj1 x2 = Inj1 x3).
Apply unknownprop_3848cfb1fd522cb609408da39f227a9c05924a24919f21041d0880590b824ef5 with λ x3 . In x3 x1, λ x3 . Inj1 x2 = Inj1 x3, In x2 x1 leaving 2 subgoals.
The subproof is completed by applying H1.
Let x3 of type ι be given.
Assume H2: In x3 x1.
Assume H3: Inj1 x2 = Inj1 x3.
Apply unknownprop_da82029798a5f9bbb993db5eff220fcf8fcae5b2fe3af56ebe878a8d8aeef3c3 with x2, x3, λ x4 x5 . In x5 x1 leaving 2 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H2.