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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι → ι → ι be given.
Assume H0: explicit_Group x0 x1.
Let x2 of type ι be given.
Assume H1: prim1 x2 x0.
Apply explicit_Group_identity_prop with x0, x1, x1 (explicit_Group_identity x0 x1) x2 = x2 leaving 2 subgoals.
The subproof is completed by applying H0.
Assume H2: prim1 (explicit_Group_identity x0 x1) x0.
Assume H3: and (∀ x3 . prim1 x3 x0 ⟶ and (x1 (explicit_Group_identity x0 x1) x3 = x3) (x1 x3 (explicit_Group_identity x0 x1) = x3)) (∀ x3 . prim1 x3 x0 ⟶ ∃ x4 . and (prim1 x4 x0) (and (x1 x3 x4 = explicit_Group_identity x0 x1) (x1 x4 x3 = explicit_Group_identity x0 x1))).
Apply H3 with x1 (explicit_Group_identity x0 x1) x2 = x2.
Assume H4: ∀ x3 . prim1 x3 x0 ⟶ and (x1 (explicit_Group_identity x0 x1) x3 = x3) (x1 x3 (explicit_Group_identity x0 x1) = x3).
Assume H5: ∀ x3 . prim1 x3 x0 ⟶ ∃ x4 . and (prim1 x4 x0) (and (x1 x3 x4 = explicit_Group_identity x0 x1) (x1 x4 x3 = explicit_Group_identity x0 x1)).
Apply H4 with x2, x1 (explicit_Group_identity x0 x1) x2 = x2 leaving 2 subgoals.
The subproof is completed by applying H1.
Assume H6: x1 (explicit_Group_identity x0 x1) x2 = x2.
Assume H7: x1 x2 (explicit_Group_identity x0 x1) = x2.
The subproof is completed by applying H6.
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