Termination Proof using AProVE

Term Rewriting System R:
[x, y, n, m]
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

EQ(s(x), s(y)) -> EQ(x, y)
LE(s(x), s(y)) -> LE(x, y)
APP(add(n, x), y) -> APP(x, y)
MIN(add(n, add(m, x))) -> IF_MIN(le(n, m), add(n, add(m, x)))
MIN(add(n, add(m, x))) -> LE(n, m)
IF_MIN(true, add(n, add(m, x))) -> MIN(add(n, x))
IF_MIN(false, add(n, add(m, x))) -> MIN(add(m, x))
RM(n, add(m, x)) -> IF_RM(eq(n, m), n, add(m, x))
RM(n, add(m, x)) -> EQ(n, m)
IF_RM(true, n, add(m, x)) -> RM(n, x)
IF_RM(false, n, add(m, x)) -> RM(n, x)
MINSORT(add(n, x), y) -> IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
MINSORT(add(n, x), y) -> EQ(n, min(add(n, x)))
MINSORT(add(n, x), y) -> MIN(add(n, x))
IF_MINSORT(true, add(n, x), y) -> MINSORT(app(rm(n, x), y), nil)
IF_MINSORT(true, add(n, x), y) -> APP(rm(n, x), y)
IF_MINSORT(true, add(n, x), y) -> RM(n, x)
IF_MINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))

Furthermore, R contains six SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pair:

EQ(s(x), s(y)) -> EQ(x, y)

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

EQ(s(x), s(y)) -> EQ(x, y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(EQ(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pair:

APP(add(n, x), y) -> APP(x, y)

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(add(n, x), y) -> APP(x, y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(APP(x₁, x₂)) = x₁

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polynomial Ordering

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pairs:

IF_RM(false, n, add(m, x)) -> RM(n, x)
IF_RM(true, n, add(m, x)) -> RM(n, x)
RM(n, add(m, x)) -> IF_RM(eq(n, m), n, add(m, x))

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_RM(false, n, add(m, x)) -> RM(n, x)
IF_RM(true, n, add(m, x)) -> RM(n, x)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_RM(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(false) = 0

POL(true) = 0

POL(s(x₁)) = 0

POL(RM(x₁, x₂)) = x₂

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

Dependency Pair:

RM(n, add(m, x)) -> IF_RM(eq(n, m), n, add(m, x))

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

Dependency Pairs:

IF_MIN(false, add(n, add(m, x))) -> MIN(add(m, x))
IF_MIN(true, add(n, add(m, x))) -> MIN(add(n, x))
MIN(add(n, add(m, x))) -> IF_MIN(le(n, m), add(n, add(m, x)))

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_MIN(false, add(n, add(m, x))) -> MIN(add(m, x))
IF_MIN(true, add(n, add(m, x))) -> MIN(add(n, x))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(false) = 0

POL(MIN(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 0

POL(IF_MIN(x₁, x₂)) = x₂

POL(le(x₁, x₂)) = 0

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 6

         ↳Polo

Dependency Pair:

MIN(add(n, add(m, x))) -> IF_MIN(le(n, m), add(n, add(m, x)))

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polynomial Ordering

Dependency Pairs:

IF_MINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))
IF_MINSORT(true, add(n, x), y) -> MINSORT(app(rm(n, x), y), nil)
MINSORT(add(n, x), y) -> IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MINSORT(true, add(n, x), y) -> MINSORT(app(rm(n, x), y), nil)

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(false) = 0

POL(if_rm(x₁, x₂, x₃)) = x₃

POL(true) = 0

POL(rm(x₁, x₂)) = x₂

POL(MINSORT(x₁, x₂)) = x₁ + x₂

POL(if_min(x₁, x₂)) = 0

POL(add(x₁, x₂)) = 1 + x₂

POL(IF_MINSORT(x₁, x₂, x₃)) = x₂ + x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(nil) = 0

POL(min(x₁)) = 0

POL(s(x₁)) = 0

POL(le(x₁, x₂)) = 0

POL(app(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

           →DP Problem 12

             ↳Polynomial Ordering

Dependency Pairs:

IF_MINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))
MINSORT(add(n, x), y) -> IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINSORT(x₁, x₂, x₃)) = x₂

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(false) = 0

POL(true) = 0

POL(min(x₁)) = 0

POL(nil) = 0

POL(s(x₁)) = 0

POL(MINSORT(x₁, x₂)) = x₁

POL(le(x₁, x₂)) = 0

POL(if_min(x₁, x₂)) = 0

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

           →DP Problem 12

             ↳Polo

...

               →DP Problem 13

                 ↳Dependency Graph

Dependency Pair:

MINSORT(add(n, x), y) -> IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)

Rules:

eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) -> min(add(n, x))
if_min(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) -> rm(n, x)
if_rm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) -> minsort(x, add(n, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(APP(x₁, x₂))	= x₁
POL(add(x₁, x₂))	= 1 + x₂

POL(IF_RM(x₁, x₂, x₃))	= x₃
POL(eq(x₁, x₂))	= 0
POL(0)	= 0
POL(false)	= 0
POL(true)	= 0
POL(s(x₁))	= 0
POL(RM(x₁, x₂))	= x₂
POL(add(x₁, x₂))	= 1 + x₂

POL(0)	= 0
POL(false)	= 0
POL(MIN(x₁))	= x₁
POL(true)	= 0
POL(s(x₁))	= 0
POL(IF_MIN(x₁, x₂))	= x₂
POL(le(x₁, x₂))	= 0
POL(add(x₁, x₂))	= 1 + x₂

POL(IF_MINSORT(x₁, x₂, x₃))	= x₂
POL(eq(x₁, x₂))	= 0
POL(0)	= 0
POL(false)	= 0
POL(true)	= 0
POL(min(x₁))	= 0
POL(nil)	= 0
POL(s(x₁))	= 0
POL(MINSORT(x₁, x₂))	= x₁
POL(le(x₁, x₂))	= 0
POL(if_min(x₁, x₂))	= 0
POL(add(x₁, x₂))	= 1 + x₂