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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 ιιι be given.
Let x6 of type ιιι be given.
Assume H0: ∀ x7 . x7x0∀ x8 . x8x0x3 x7 x8 = x5 x7 x8.
Assume H1: ∀ x7 . x7x0∀ x8 . x8x0x4 x7 x8 = x6 x7 x8.
Apply explicit_Ring_with_id_E with x0, x1, x2, x3, x4, explicit_Ring_with_id x0 x1 x2 x5 x6.
Assume H2: explicit_Ring_with_id x0 x1 x2 x3 x4.
Assume H3: ∀ x7 . x7x0∀ x8 . x8x0x3 x7 x8x0.
Assume H4: ∀ x7 . x7x0∀ x8 . x8x0∀ x9 . x9x0x3 x7 (x3 x8 x9) = x3 (x3 x7 x8) x9.
Assume H5: ∀ x7 . x7x0∀ x8 . x8x0x3 x7 x8 = x3 x8 x7.
Assume H6: x1x0.
Assume H7: ∀ x7 . x7x0x3 x1 x7 = x7.
Assume H8: ∀ x7 . x7x0∃ x8 . and (x8x0) (x3 x7 x8 = x1).
Assume H9: ∀ x7 . x7x0∀ x8 . x8x0x4 x7 x8x0.
Assume H10: ∀ x7 . x7x0∀ x8 . x8x0∀ x9 . x9x0x4 x7 (x4 x8 x9) = x4 (x4 x7 x8) x9.
Assume H11: x2x0.
Assume H12: x2 = x1∀ x7 : ο . x7.
Assume H13: ∀ x7 . x7x0x4 x2 x7 = x7.
Assume H14: ∀ x7 . x7x0x4 x7 x2 = x7.
Assume H15: ∀ x7 . x7x0∀ x8 . x8x0∀ x9 . x9x0x4 x7 (x3 x8 x9) = x3 (x4 x7 x8) (x4 x7 x9).
Assume H16: ∀ x7 . x7x0∀ x8 . x8x0∀ x9 . x9x0x4 (x3 x7 x8) x9 = x3 (x4 x7 x9) (x4 x8 x9).
Apply explicit_Ring_with_id_I with x0, x1, x2, x5, x6 leaving 14 subgoals.
Let x7 of type ι be given.
Assume H17: x7x0.
Let x8 of type ι be given.
Assume H18: x8x0.
Apply H0 with x7, x8, λ x9 x10 . x9x0 leaving 3 subgoals.
The subproof is completed by applying H17.
The subproof is completed by applying H18.
Apply H3 with x7, x8 leaving 2 subgoals.
The subproof is completed by applying H17.
The subproof is completed by applying H18.
Let x7 of type ι be given.
Assume H17: x7x0.
Let x8 of type ι be given.
Assume H18: x8....
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