Let x0 of type ι be given.

Let x1 of type ι → ι → ι be given.

Apply explicit_Group_identity_prop with x0, x1, ∀ x2 . x2 ∈ x0 ⟶ ∃ x3 . and (x3 ∈ x0) (and (x1 x2 x3 = explicit_Group_identity x0 x1) (x1 x3 x2 = explicit_Group_identity x0 x1)) leaving 2 subgoals.

The subproof is completed by applying H0.

Assume H1: explicit_Group_identity x0 x1 ∈ x0.

Assume H2: and (∀ x2 . x2 ∈ x0 ⟶ and (x1 (explicit_Group_identity x0 x1) x2 = x2) (x1 x2 (explicit_Group_identity x0 x1) = x2)) (∀ x2 . x2 ∈ x0 ⟶ ∃ x3 . and (x3 ∈ x0) (and (x1 x2 x3 = explicit_Group_identity x0 x1) (x1 x3 x2 = explicit_Group_identity x0 x1))).

Apply H2 with ∀ x2 . x2 ∈ x0 ⟶ ∃ x3 . and (x3 ∈ x0) (and (x1 x2 x3 = explicit_Group_identity x0 x1) (x1 x3 x2 = explicit_Group_identity x0 x1)).

Assume H3: ∀ x2 . x2 ∈ x0 ⟶ and (x1 (explicit_Group_identity x0 x1) x2 = x2) (x1 x2 (explicit_Group_identity x0 x1) = x2).

Assume H4: ∀ x2 . x2 ∈ x0 ⟶ ∃ x3 . and (x3 ∈ x0) (and (x1 x2 x3 = explicit_Group_identity x0 x1) (x1 x3 x2 = explicit_Group_identity x0 x1)).

The subproof is completed by applying H4.

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Proofgold Proof