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Proofgold Proof
pf
Let x0 of type
ι
be given.
Let x1 of type
ι
be given.
Assume H0:
SNo
x0
.
Assume H1:
∀ x2 .
x2
∈
SNoS_
(
SNoLev
x0
)
⟶
and
(
and
(
and
(
SNo
(
minus_SNo
x2
)
)
(
∀ x3 .
x3
∈
SNoL
x2
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x3
)
)
)
(
∀ x3 .
x3
∈
SNoR
x2
⟶
SNoLt
(
minus_SNo
x3
)
(
minus_SNo
x2
)
)
)
(
SNoCutP
(
prim5
(
SNoR
x2
)
minus_SNo
)
(
prim5
(
SNoL
x2
)
minus_SNo
)
)
.
Assume H2:
SNoLev
x1
∈
SNoLev
x0
.
Assume H3:
x1
∈
SNoS_
(
SNoLev
x0
)
.
Apply H1 with
x1
,
and
(
and
(
SNo
(
minus_SNo
x1
)
)
(
∀ x2 .
x2
∈
SNoL
x1
⟶
SNoLt
(
minus_SNo
x1
)
(
minus_SNo
x2
)
)
)
(
∀ x2 .
x2
∈
SNoR
x1
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x1
)
)
leaving 2 subgoals.
The subproof is completed by applying H3.
Assume H4:
and
(
and
(
SNo
(
minus_SNo
x1
)
)
(
∀ x2 .
x2
∈
SNoL
x1
⟶
SNoLt
(
minus_SNo
x1
)
(
minus_SNo
x2
)
)
)
(
∀ x2 .
x2
∈
SNoR
x1
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x1
)
)
.
Apply H4 with
SNoCutP
(
prim5
(
SNoR
x1
)
minus_SNo
)
(
prim5
(
SNoL
x1
)
minus_SNo
)
⟶
and
(
and
(
SNo
(
minus_SNo
x1
)
)
(
∀ x2 .
x2
∈
SNoL
x1
⟶
SNoLt
(
minus_SNo
x1
)
(
minus_SNo
x2
)
)
)
(
∀ x2 .
x2
∈
SNoR
x1
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x1
)
)
.
Assume H5:
and
(
SNo
(
minus_SNo
x1
)
)
(
∀ x2 .
x2
∈
SNoL
x1
⟶
SNoLt
(
minus_SNo
x1
)
(
minus_SNo
x2
)
)
.
Apply H5 with
(
∀ x2 .
x2
∈
SNoR
x1
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x1
)
)
⟶
SNoCutP
(
prim5
(
SNoR
x1
)
minus_SNo
)
(
prim5
(
SNoL
x1
)
minus_SNo
)
⟶
and
(
and
(
SNo
(
minus_SNo
x1
)
)
(
∀ x2 .
x2
∈
SNoL
x1
⟶
SNoLt
(
minus_SNo
x1
)
(
minus_SNo
x2
)
)
)
(
∀ x2 .
x2
∈
SNoR
x1
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x1
)
)
.
Assume H6:
SNo
(
minus_SNo
x1
)
.
Assume H7:
∀ x2 .
x2
∈
SNoL
x1
⟶
SNoLt
(
minus_SNo
x1
)
(
minus_SNo
x2
)
.
Assume H8:
∀ x2 .
x2
∈
SNoR
x1
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x1
)
.
Assume H9:
SNoCutP
(
prim5
(
SNoR
x1
)
minus_SNo
)
(
prim5
(
SNoL
x1
)
minus_SNo
)
.
Apply and3I with
SNo
(
minus_SNo
x1
)
,
∀ x2 .
x2
∈
SNoL
x1
⟶
SNoLt
(
minus_SNo
x1
)
(
minus_SNo
x2
)
,
∀ x2 .
x2
∈
SNoR
x1
⟶
SNoLt
(
minus_SNo
x2
)
(
minus_SNo
x1
)
leaving 3 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
The subproof is completed by applying H8.
■