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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
x2
(
explicit_Group_identity
x0
x1
)
=
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
x2
(
explicit_Group_identity
x0
x1
)
=
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
x2
(
explicit_Group_identity
x0
x1
)
=
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 H7.
■