Search for blocks/addresses/...

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: 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: explicit_Group_identity x0 x1 ∈ x0.
Assume H3: and (∀ x3 . x3 ∈ x0 ⟶ and (x1 (explicit_Group_identity x0 x1) x3 = x3) (x1 x3 (explicit_Group_identity x0 x1) = x3)) (∀ x3 . x3 ∈ x0 ⟶ ∃ x4 . and (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 . x3 ∈ x0 ⟶ and (x1 (explicit_Group_identity x0 x1) x3 = x3) (x1 x3 (explicit_Group_identity x0 x1) = x3).
Assume H5: ∀ x3 . x3 ∈ x0 ⟶ ∃ x4 . and (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.
■