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Proofgold Proof

pf
Claim L0: ∀ x0 . 74e69.. x0∀ x1 x2 . x0 = a3eb9.. x1 x2and (74e69.. x1) (74e69.. x2)
Apply unknownprop_9eb52b34dc5ec3159c14c43cf9730f7233c1ae17392d54fc91e463c1c0c6ba1d with λ x0 . ∀ x1 x2 . x0 = a3eb9.. x1 x2and (74e69.. x1) (74e69.. x2) leaving 3 subgoals.
Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H0: 5e331.. = a3eb9.. x0 x1.
Apply FalseE with and (74e69.. x0) (74e69.. x1).
Apply unknownprop_112da44b75efcdc251767c8f8619fc43c6b883afd5ac71e5854bf36b62571f0c with x0, x1.
The subproof is completed by applying H0.
Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H0: 74e69.. x0.
Assume H1: ∀ x2 x3 . x0 = a3eb9.. x2 x3and (74e69.. x2) (74e69.. x3).
Assume H2: 74e69.. x1.
Assume H3: ∀ x2 x3 . x1 = a3eb9.. x2 x3and (74e69.. x2) (74e69.. x3).
Let x2 of type ι be given.
Let x3 of type ι be given.
Assume H4: a3eb9.. x0 x1 = a3eb9.. x2 x3.
Apply unknownprop_1f81fbea8647063ecd1729d03bf27a5327e4a018cc4ad3e99f64b38027811c93 with x0, x1, x2, x3, and (74e69.. x2) (74e69.. x3) leaving 2 subgoals.
The subproof is completed by applying H4.
Assume H5: x0 = x2.
Assume H6: x1 = x3.
Apply andI with 74e69.. x2, 74e69.. x3 leaving 2 subgoals.
Apply H5 with λ x4 x5 . 74e69.. x4.
The subproof is completed by applying H0.
Apply H6 with λ x4 x5 . 74e69.. x4.
The subproof is completed by applying H2.
Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H0: 74e69.. x0.
Assume H1: ∀ x2 x3 . x0 = a3eb9.. x2 x3and (74e69.. x2) (74e69.. x3).
Assume H2: 74e69.. x1.
Assume H3: ∀ x2 x3 . x1 = a3eb9.. x2 x3and (74e69.. x2) (74e69.. x3).
Let x2 of type ι be given.
Let x3 of type ι be given.
Assume H4: bf68c.. x0 x1 = a3eb9.. x2 x3.
Apply FalseE with and (74e69.. x2) (74e69.. x3).
Apply unknownprop_856b8bfadc699ad231b4ff87ec1e03728d77f11011b06095d14ff77ccdcd01b1 with x2, x3, x0, x1.
Let x4 of type ιιο be given.
The subproof is completed by applying H4 with λ x5 x6 . x4 x6 x5.
Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H1: 74e69.. (a3eb9.. x0 x1).
Apply L0 with a3eb9.. x0 x1, x0, x1 leaving 2 subgoals.
The subproof is completed by applying H1.
Let x2 of type ιιο be given.
Assume H2: x2 (a3eb9.. x0 x1) (a3eb9.. x0 x1).
The subproof is completed by applying H2.