Search for blocks/addresses/...
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.
Assume H0:
prim1
x3
(
0fc90..
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
(
λ x4 .
If_i
(
x4
=
4a7ef..
)
x0
(
If_i
(
x4
=
4ae4a..
4a7ef..
)
x1
x2
)
)
)
.
Claim L1:
∃ x4 .
and
(
prim1
x4
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
(
∃ x5 .
and
(
prim1
x5
(
If_i
(
x4
=
4a7ef..
)
x0
(
If_i
(
x4
=
4ae4a..
4a7ef..
)
x1
x2
)
)
)
(
x3
=
aae7a..
x4
x5
)
)
Apply unknownprop_d4dc73f3cbfe4c22363272ac418d035b8e77c49433b0870b96bee8fa9c46bfcf with
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
,
λ x4 .
If_i
(
x4
=
4a7ef..
)
x0
(
If_i
(
x4
=
4ae4a..
4a7ef..
)
x1
x2
)
,
x3
.
The subproof is completed by applying H0.
Apply exandE_i with
λ x4 .
prim1
x4
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
,
λ x4 .
∃ x5 .
and
(
prim1
x5
(
If_i
(
x4
=
4a7ef..
)
x0
(
If_i
(
x4
=
4ae4a..
4a7ef..
)
x1
x2
)
)
)
(
x3
=
aae7a..
x4
x5
)
,
∃ x4 .
and
(
prim1
x4
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
(
∃ x5 .
x3
=
aae7a..
x4
x5
)
leaving 2 subgoals.
The subproof is completed by applying L1.
Let x4 of type
ι
be given.
Assume H2:
prim1
x4
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
.
Assume H3:
∃ x5 .
and
(
prim1
x5
(
If_i
(
x4
=
4a7ef..
)
x0
(
If_i
(
x4
=
4ae4a..
4a7ef..
)
x1
x2
)
)
)
(
x3
=
aae7a..
x4
x5
)
.
Apply exandE_i with
λ x5 .
prim1
x5
(
If_i
(
x4
=
4a7ef..
)
x0
(
If_i
(
x4
=
4ae4a..
4a7ef..
)
x1
x2
)
)
,
λ x5 .
x3
=
aae7a..
x4
x5
,
∃ x5 .
and
(
prim1
x5
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
(
∃ x6 .
x3
=
aae7a..
x5
x6
)
leaving 2 subgoals.
The subproof is completed by applying H3.
Let x5 of type
ι
be given.
Assume H4:
prim1
x5
(
If_i
(
x4
=
4a7ef..
)
x0
(
If_i
(
x4
=
4ae4a..
4a7ef..
)
x1
x2
)
)
.
Assume H5:
x3
=
aae7a..
x4
x5
.
Let x6 of type
ο
be given.
Assume H6:
∀ x7 .
and
(
prim1
x7
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
)
(
∃ x8 .
x3
=
aae7a..
x7
x8
)
⟶
x6
.
Apply H6 with
x4
.
Apply andI with
prim1
x4
(
4ae4a..
(
4ae4a..
(
4ae4a..
4a7ef..
)
)
)
,
∃ x7 .
x3
=
aae7a..
x4
x7
leaving 2 subgoals.
The subproof is completed by applying H2.
Let x7 of type
ο
be given.
Assume H7:
∀ x8 .
x3
=
aae7a..
x4
x8
⟶
x7
.
Apply H7 with
x5
.
The subproof is completed by applying H5.
■