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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.
Let x3 of type ι be given.
Apply unknownprop_2c926b240fc658005337215abfdc8124a6f6eea17ba9f4df80d254bab3845972 with λ x4 x5 : ι → ι → ο . x5 4 (lam 4 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 (If_i (x6 = 2) x2 x3)))).
Let x4 of type ι be given.
Assume H0: In x4 (lam 4 (λ x5 . If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3)))).
Claim L1: ∃ x5 . and (In x5 4) (∃ x6 . and (In x6 (If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3)))) (x4 = setsum x5 x6))
Apply unknownprop_f25818182af6b093121a8b5d43847162c8ea91396e524cca02557613a430a57a with 4, λ x5 . If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3)), x4.
The subproof is completed by applying H0.
Apply unknownprop_3848cfb1fd522cb609408da39f227a9c05924a24919f21041d0880590b824ef5 with λ x5 . In x5 4, λ x5 . ∃ x6 . and (In x6 (If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3)))) (x4 = setsum x5 x6), ∃ x5 . and (In x5 4) (∃ x6 . x4 = setsum x5 x6) leaving 2 subgoals.
The subproof is completed by applying L1.
Let x5 of type ι be given.
Assume H2: In x5 4.
Assume H3: ∃ x6 . and (In x6 (If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3)))) (x4 = setsum x5 x6).
Apply unknownprop_3848cfb1fd522cb609408da39f227a9c05924a24919f21041d0880590b824ef5 with λ x6 . In x6 (If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3))), λ x6 . x4 = setsum x5 x6, ∃ x6 . and (In x6 4) (∃ x7 . x4 = setsum x6 x7) leaving 2 subgoals.
The subproof is completed by applying H3.
Let x6 of type ι be given.
Assume H4: In x6 (If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3))).
Assume H5: x4 = setsum x5 x6.
Let x7 of type ο be given.
Assume H6: ∀ x8 . and (In x8 4) (∃ x9 . x4 = setsum x8 x9)x7.
Apply H6 with x5.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with In x5 4, ∃ x8 . x4 = setsum x5 x8 leaving 2 subgoals.
The subproof is completed by applying H2.
Let x8 of type ο be given.
Assume H7: ∀ x9 . x4 = setsum x5 x9x8.
Apply H7 with x6.
The subproof is completed by applying H5.