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Proofgold Proof
pf
Let x0 of type
ι
be given.
Let x1 of type
ι
be given.
Let x2 of type
ι
be given.
Let x3 of type
ι
be given.
Let x4 of type
ι
be given.
Assume H0:
prim1
x4
(
0fc90..
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
(
λ x5 .
If_i
(
x5
=
4a7ef..
)
x0
(
If_i
(
x5
=
4ae4a..
4a7ef..
)
x1
(
If_i
(
x5
=
4ae4a..
(
4ae4a..
4a7ef..
)
)
x2
x3
)
)
)
)
.
Claim L1:
∃ x5 .
and
(
prim1
x5
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
)
(
∃ x6 .
and
(
prim1
x6
(
If_i
(
x5
=
4a7ef..
)
x0
(
If_i
(
x5
=
4ae4a..
4a7ef..
)
x1
(
If_i
(
x5
=
4ae4a..
(
4ae4a..
4a7ef..
)
)
...
...
)
)
)
)
...
)
...
Apply exandE_i with
λ x5 .
prim1
x5
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
,
λ x5 .
∃ x6 .
and
(
prim1
x6
(
If_i
(
x5
=
4a7ef..
)
x0
(
If_i
(
x5
=
4ae4a..
4a7ef..
)
x1
(
If_i
(
x5
=
4ae4a..
(
4ae4a..
4a7ef..
)
)
x2
x3
)
)
)
)
(
x4
=
aae7a..
x5
x6
)
,
∃ x5 .
and
(
prim1
x5
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
)
(
∃ x6 .
x4
=
aae7a..
x5
x6
)
leaving 2 subgoals.
The subproof is completed by applying L1.
Let x5 of type
ι
be given.
Assume H2:
prim1
x5
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
.
Assume H3:
∃ x6 .
and
(
prim1
x6
(
If_i
(
x5
=
4a7ef..
)
x0
(
If_i
(
x5
=
4ae4a..
4a7ef..
)
x1
(
If_i
(
x5
=
4ae4a..
(
4ae4a..
4a7ef..
)
)
x2
x3
)
)
)
)
(
x4
=
aae7a..
x5
x6
)
.
Apply exandE_i with
λ x6 .
prim1
x6
(
If_i
(
x5
=
4a7ef..
)
x0
(
If_i
(
x5
=
4ae4a..
4a7ef..
)
x1
(
If_i
(
x5
=
4ae4a..
(
4ae4a..
4a7ef..
)
)
x2
x3
)
)
)
,
λ x6 .
x4
=
aae7a..
x5
x6
,
∃ x6 .
and
(
prim1
x6
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
)
(
∃ x7 .
x4
=
aae7a..
x6
x7
)
leaving 2 subgoals.
The subproof is completed by applying H3.
Let x6 of type
ι
be given.
Assume H4:
prim1
x6
(
If_i
(
x5
=
4a7ef..
)
x0
(
If_i
(
x5
=
4ae4a..
4a7ef..
)
x1
(
If_i
(
x5
=
4ae4a..
(
4ae4a..
4a7ef..
)
)
x2
x3
)
)
)
.
Assume H5:
x4
=
aae7a..
x5
x6
.
Let x7 of type
ο
be given.
Assume H6:
∀ x8 .
and
(
prim1
x8
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
)
(
∃ x9 .
x4
=
aae7a..
x8
x9
)
⟶
x7
.
Apply H6 with
x5
.
Apply andI with
prim1
x5
(
4ae4a..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
,
∃ x8 .
x4
=
aae7a..
x5
x8
leaving 2 subgoals.
The subproof is completed by applying H2.
Let x8 of type
ο
be given.
Assume H7:
∀ x9 .
x4
=
aae7a..
x5
x9
⟶
x8
.
Apply H7 with
x6
.
The subproof is completed by applying H5.
■