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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 ιιι be given.
Let x3 of type ιιι be given.
Let x4 of type ο be given.
Assume H0: explicit_Ring x0 x1 x2 x3(∀ x5 . x5x0∀ x6 . x6x0x2 x5 x6x0)(∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x2 x5 (x2 x6 x7) = x2 (x2 x5 x6) x7)(∀ x5 . x5x0∀ x6 . x6x0x2 x5 x6 = x2 x6 x5)x1x0(∀ x5 . x5x0x2 x1 x5 = x5)(∀ x5 . x5x0∃ x6 . and (x6x0) (x2 x5 x6 = x1))(∀ x5 . x5x0∀ x6 . x6x0x3 x5 x6x0)(∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x3 x5 (x3 x6 x7) = x3 (x3 x5 x6) x7)(∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x3 x5 (x2 x6 x7) = x2 (x3 x5 x6) (x3 x5 x7))(∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x3 (x2 x5 x6) x7 = x2 (x3 x5 x7) (x3 x6 x7))x4.
Assume H1: explicit_Ring x0 x1 x2 x3.
Apply and4E with and (and (and (and (and (and (∀ x5 . x5x0∀ x6 . x6x0x2 x5 x6x0) (∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x2 x5 (x2 x6 x7) = x2 (x2 x5 x6) x7)) (∀ x5 . x5x0∀ x6 . x6x0x2 x5 x6 = x2 x6 x5)) (x1x0)) (∀ x5 . x5x0x2 x1 x5 = x5)) (∀ x5 . x5x0∃ x6 . and (x6x0) (x2 x5 x6 = x1))) (∀ x5 . x5x0∀ x6 . x6x0x3 x5 x6x0), ∀ x5 . ...∀ x6 . ...∀ x7 . ..., ..., ..., ... leaving 2 subgoals.
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