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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 ιιι be given.
Let x4 of type ιιι be given.
Let x5 of type ιιο be given.
Apply explicit_OrderedField_E with x0, x1, x2, x3, x4, x5, ∀ x6 . x6x0x5 x6 x6.
Assume H0: explicit_OrderedField x0 x1 x2 x3 x4 x5.
Assume H1: explicit_Field x0 x1 x2 x3 x4.
Assume H2: ∀ x6 . x6x0∀ x7 . x7x0∀ x8 . x8x0x5 x6 x7x5 x7 x8x5 x6 x8.
Assume H3: ∀ x6 . x6x0∀ x7 . x7x0iff (and (x5 x6 x7) (x5 x7 x6)) (x6 = x7).
Assume H4: ∀ x6 . x6x0∀ x7 . x7x0or (x5 x6 x7) (x5 x7 x6).
Assume H5: ∀ x6 . x6x0∀ x7 . x7x0∀ x8 . x8x0x5 x6 x7x5 (x3 x6 x8) (x3 x7 x8).
Assume H6: ∀ x6 . x6x0∀ x7 . x7x0x5 x1 x6x5 x1 x7x5 x1 (x4 x6 x7).
Let x6 of type ι be given.
Assume H7: x6x0.
Apply H3 with x6, x6, x5 x6 x6 leaving 3 subgoals.
The subproof is completed by applying H7.
The subproof is completed by applying H7.
Assume H8: and (x5 x6 x6) (x5 x6 x6)x6 = x6.
Assume H9: x6 = x6and (x5 x6 x6) (x5 x6 x6).
Apply H9 with x5 x6 x6 leaving 2 subgoals.
Let x7 of type ιιο be given.
Assume H10: x7 x6 x6.
The subproof is completed by applying H10.
Assume H10: x5 x6 x6.
Assume H11: x5 x6 x6.
The subproof is completed by applying H10.