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Proofgold Proof
pf
Let x0 of type
ι
be given.
Let x1 of type
ι
be given.
Assume H0:
Subq
x1
x0
.
Let x2 of type
ι
be given.
Assume H1:
prim1
x2
(
1216a..
(
b5c9f..
x0
x0
)
(
λ x3 .
and
(
bij
x0
x0
(
λ x4 .
f482f..
x3
x4
)
)
(
∀ x4 .
prim1
x4
x1
⟶
f482f..
x3
x4
=
x4
)
)
)
.
Let x3 of type
ι
be given.
Assume H2:
prim1
x3
(
1216a..
(
b5c9f..
x0
x0
)
(
λ x4 .
and
(
bij
x0
x0
(
λ x5 .
f482f..
x4
x5
)
)
(
∀ x5 .
prim1
x5
x1
⟶
f482f..
x4
x5
=
x5
)
)
)
.
Apply unknownprop_e4362c04e65a765de9cf61045b78be0adc0f9e51a17754420e1088df0891ff67 with
b5c9f..
x0
x0
,
λ x4 .
and
(
bij
x0
x0
(
λ x5 .
f482f..
x4
x5
)
)
(
∀ x5 .
prim1
x5
x1
⟶
f482f..
x4
x5
=
x5
)
,
x2
,
prim1
(
(
λ x4 x5 .
0fc90..
x0
(
λ x6 .
f482f..
x5
(
f482f..
x4
x6
)
)
)
x2
x3
)
(
1216a..
(
b5c9f..
x0
x0
)
(
λ x4 .
and
(
bij
x0
x0
(
λ x5 .
f482f..
x4
x5
)
)
(
∀ x5 .
prim1
x5
x1
⟶
f482f..
x4
x5
=
x5
)
)
)
leaving 2 subgoals.
The subproof is completed by applying H1.
Assume H3:
prim1
x2
(
b5c9f..
x0
x0
)
.
Assume H4:
and
(
bij
x0
x0
(
f482f..
x2
)
)
(
∀ x4 .
prim1
x4
x1
⟶
f482f..
x2
x4
=
x4
)
.
Apply H4 with
prim1
(
(
λ x4 x5 .
0fc90..
x0
(
λ x6 .
f482f..
x5
(
f482f..
x4
x6
)
)
)
x2
x3
)
(
1216a..
(
b5c9f..
x0
x0
)
(
λ x4 .
and
(
bij
x0
x0
(
λ x5 .
f482f..
x4
x5
)
)
(
∀ x5 .
prim1
x5
x1
⟶
f482f..
x4
x5
=
x5
)
)
)
.
Assume H5:
bij
x0
x0
(
λ x4 .
f482f..
x2
x4
)
.
Assume H6:
∀ x4 .
prim1
x4
x1
⟶
f482f..
x2
x4
=
x4
.
Claim L7:
...
...
Apply unknownprop_e4362c04e65a765de9cf61045b78be0adc0f9e51a17754420e1088df0891ff67 with
b5c9f..
x0
x0
,
λ x4 .
and
(
bij
x0
x0
(
λ x5 .
f482f..
x4
x5
)
)
(
∀ x5 .
prim1
x5
x1
⟶
f482f..
x4
x5
=
x5
)
,
x3
,
prim1
(
(
λ x4 x5 .
0fc90..
x0
(
λ x6 .
f482f..
x5
(
f482f..
x4
x6
)
)
)
x2
x3
)
(
1216a..
(
b5c9f..
x0
x0
)
(
λ x4 .
and
(
bij
x0
x0
(
λ x5 .
f482f..
x4
x5
)
)
(
∀ x5 .
prim1
x5
x1
⟶
f482f..
x4
x5
=
x5
)
)
)
leaving 2 subgoals.
The subproof is completed by applying H2.
Assume H8:
prim1
x3
(
b5c9f..
x0
x0
)
.
Assume H9:
and
(
bij
x0
x0
(
f482f..
x3
)
)
(
∀ x4 .
prim1
...
...
⟶
f482f..
x3
x4
=
x4
)
.
...
■