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Proofgold Proof

pf
Let x0 of type ι(ιο) → ο be given.
Let x1 of type ι(ιο) → ο be given.
Assume H0: ed32f.. x0 x1.
Claim L1: ...
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Apply ordinal_ind with λ x2 . or (∃ x3 : ι → ο . 078b6.. x0 x1 x2 x3) (∃ x3 . and (prim1 x3 x2) (∃ x4 : ι → ο . a59df.. x0 x1 x3 x4)).
Let x2 of type ι be given.
Assume H2: ordinal x2.
Apply H2 with (∀ x3 . prim1 x3 x2or (∃ x4 : ι → ο . 078b6.. x0 x1 x3 x4) (∃ x4 . and (prim1 x4 x3) (∃ x5 : ι → ο . a59df.. x0 x1 x4 x5)))or (∃ x3 : ι → ο . 078b6.. x0 x1 x2 x3) (∃ x3 . and (prim1 x3 x2) (∃ x4 : ι → ο . a59df.. x0 x1 x3 x4)).
Assume H3: TransSet x2.
Assume H4: ∀ x3 . prim1 x3 x2TransSet x3.
Assume H5: ∀ x3 . prim1 x3 x2or (∃ x4 : ι → ο . 078b6.. x0 x1 x3 x4) (∃ x4 . and (prim1 x4 x3) (∃ x5 : ι → ο . a59df.. x0 x1 x4 x5)).
Apply dneg with or (∃ x3 : ι → ο . 078b6.. x0 x1 x2 x3) (∃ x3 . and (prim1 x3 x2) (∃ x4 : ι → ο . a59df.. x0 x1 x3 x4)).
Assume H6: not (or (∃ x3 : ι → ο . 078b6.. x0 x1 x2 x3) (∃ x3 . and (prim1 x3 x2) (∃ x4 : ι → ο . a59df.. x0 x1 x3 x4))).
Apply not_or_and_demorgan with ∃ x3 : ι → ο . 078b6.. x0 x1 x2 x3, ∃ x3 . and (prim1 x3 x2) (∃ x4 : ι → ο . a59df.. x0 x1 x3 x4), False leaving 2 subgoals.
The subproof is completed by applying H6.
Assume H7: not (∃ x3 : ι → ο . 078b6.. x0 x1 x2 x3).
Assume H8: not (∃ x3 . and (prim1 x3 x2) (∃ x4 : ι → ο . a59df.. x0 x1 x3 x4)).
Claim L9: ...
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Apply unknownprop_63cf8167ec21b344b97624441aeddfcf366466436f5a81c6bd27f9ec03224f29 with x2, False leaving 3 subgoals.
The subproof is completed by applying H2.
Assume H10: ∀ x3 . prim1 x3 x2prim1 (4ae4a.. x3) x2.
Claim L11: ...
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Claim L12: ...
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Apply H7.
Let x3 of type ο be given.
Assume H13: ∀ x4 : ι → ο . 078b6.. x0 x1 x2 x4x3.
Apply H13 with λ x4 . ∀ x5 : ι → ο . 8033b.. x0 x1 (4ae4a.. x4) x5x5 x4.
Apply andI with 8033b.. x0 x1 x2 (λ x4 . ∀ x5 : ι → ο . 8033b.. x0 x1 (4ae4a.. x4) x5x5 x4), ∀ x4 : ι → ο . 8033b.. x0 x1 x2 x4PNoEq_ x2 (λ x5 . ∀ x6 : ι → ο . 8033b.. x0 x1 (4ae4a.. x5) x6x6 x5) x4 leaving 2 subgoals.
Apply andI with 6f2c4.. x0 x2 (λ x4 . ∀ x5 : ι → ο . 8033b.. x0 x1 (4ae4a.. x4) x5x5 x4), dafc2.. x1 x2 ... leaving 2 subgoals.
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