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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.
Assume H0: explicit_Group x1 x2.
Assume H1: x0x1.
Assume H2: ∀ x4 . x4x1∀ x5 . x5x1x2 x4 x5 = x3 x4 x5.
Assume H3: ∀ x4 . x4x1{x2 x4 (x2 x5 (explicit_Group_inverse x1 x2 x4))|x5 ∈ x0}x0.
Let x4 of type ι be given.
Assume H4: x4x1.
Let x5 of type ι be given.
Assume H5: x5{x3 x4 (x3 x6 (explicit_Group_inverse x1 x3 x4))|x6 ∈ x0}.
Apply ReplE_impred with x0, λ x6 . x3 x4 (x3 x6 (explicit_Group_inverse x1 x3 x4)), x5, x5x0 leaving 2 subgoals.
The subproof is completed by applying H5.
Let x6 of type ι be given.
Assume H6: x6x0.
Assume H7: x5 = x3 x4 (x3 x6 (explicit_Group_inverse x1 x3 x4)).
Claim L8: ...
...
Claim L9: ...
...
Claim L10: x2 x6 (explicit_Group_inverse x1 x2 x4)x1
Apply H0 with x2 x6 (explicit_Group_inverse x1 x2 x4)x1.
Assume H10: and (∀ x7 . x7x1∀ x8 . x8x1x2 x7 x8x1) (∀ x7 . x7x1∀ x8 . x8x1∀ x9 . x9x1x2 x7 (x2 x8 x9) = x2 (x2 x7 x8) x9).
Assume H11: ∃ x7 . and (x7x1) (and (∀ x8 . ...) ...).
...
Claim L11: x5 = x2 x4 (x2 x6 (explicit_Group_inverse x1 x2 x4))
Apply H7 with λ x7 x8 . x8 = x2 x4 (x2 x6 (explicit_Group_inverse x1 x2 x4)).
Apply L8 with λ x7 x8 . x3 x4 (x3 x6 x7) = x2 x4 (x2 x6 (explicit_Group_inverse x1 x2 x4)).
Apply H2 with x6, explicit_Group_inverse x1 x2 x4, λ x7 x8 . x3 x4 x7 = x2 x4 (x2 x6 (explicit_Group_inverse x1 x2 x4)) leaving 3 subgoals.
Apply H1 with x6.
The subproof is completed by applying H6.
The subproof is completed by applying L9.
Apply H2 with x4, x2 x6 (explicit_Group_inverse x1 x2 x4), λ x7 x8 . x7 = x2 x4 (x2 x6 (explicit_Group_inverse x1 x2 x4)) leaving 3 subgoals.
The subproof is completed by applying H4.
The subproof is completed by applying L10.
Let x7 of type ιιο be given.
Assume H11: x7 (x2 x4 (x2 x6 (explicit_Group_inverse x1 x2 x4))) (x2 x4 (x2 x6 (explicit_Group_inverse x1 x2 x4))).
The subproof is completed by applying H11.
Apply H3 with x4, x5 leaving 2 subgoals.
The subproof is completed by applying H4.
Apply L11 with λ x7 x8 . x8{x2 x4 (x2 x9 (explicit_Group_inverse x1 x2 x4))|x9 ∈ x0}.
Apply ReplI with x0, λ x7 . x2 x4 (x2 x7 (explicit_Group_inverse x1 x2 x4)), x6.
The subproof is completed by applying H6.