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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
.
Apply explicit_Group_identity_prop with
x0
,
x1
,
∀ x2 .
prim1
x2
x0
⟶
∃ x3 .
and
(
prim1
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:
prim1
(
explicit_Group_identity
x0
x1
)
x0
.
Assume H2:
and
(
∀ x2 .
prim1
x2
x0
⟶
and
(
x1
(
explicit_Group_identity
x0
x1
)
x2
=
x2
)
(
x1
x2
(
explicit_Group_identity
x0
x1
)
=
x2
)
)
(
∀ x2 .
prim1
x2
x0
⟶
∃ x3 .
and
(
prim1
x3
x0
)
(
and
(
x1
x2
x3
=
explicit_Group_identity
x0
x1
)
(
x1
x3
x2
=
explicit_Group_identity
x0
x1
)
)
)
.
Apply H2 with
∀ x2 .
prim1
x2
x0
⟶
∃ x3 .
and
(
prim1
x3
x0
)
(
and
(
x1
x2
x3
=
explicit_Group_identity
x0
x1
)
(
x1
x3
x2
=
explicit_Group_identity
x0
x1
)
)
.
Assume H3:
∀ x2 .
prim1
x2
x0
⟶
and
(
x1
(
explicit_Group_identity
x0
x1
)
x2
=
x2
)
(
x1
x2
(
explicit_Group_identity
x0
x1
)
=
x2
)
.
Assume H4:
∀ x2 .
prim1
x2
x0
⟶
∃ x3 .
and
(
prim1
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.
■