Termination Proof using AProVE

Term Rewriting System R:
[m, n, x, k]
eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

EQ(s(n), s(m)) -> EQ(n, m)
LE(s(n), s(m)) -> LE(n, m)
MIN(cons(n, cons(m, x))) -> IF_MIN(le(n, m), cons(n, cons(m, x)))
MIN(cons(n, cons(m, x))) -> LE(n, m)
IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))
REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))
REPLACE(n, m, cons(k, x)) -> EQ(n, k)
IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
SORT(cons(n, x)) -> MIN(cons(n, x))
SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))
SORT(cons(n, x)) -> REPLACE(min(cons(n, x)), n, x)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

EQ(s(n), s(m)) -> EQ(n, m)

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

EQ(s(n), s(m)) -> EQ(n, m)

one new Dependency Pair is created:

EQ(s(s(n'')), s(s(m''))) -> EQ(s(n''), s(m''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

EQ(s(s(n'')), s(s(m''))) -> EQ(s(n''), s(m''))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

EQ(s(s(n'')), s(s(m''))) -> EQ(s(n''), s(m''))

one new Dependency Pair is created:

EQ(s(s(s(n''''))), s(s(s(m'''')))) -> EQ(s(s(n'''')), s(s(m'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

EQ(s(s(s(n''''))), s(s(s(m'''')))) -> EQ(s(s(n'''')), s(s(m'''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

EQ(s(s(s(n''''))), s(s(s(m'''')))) -> EQ(s(s(n'''')), s(s(m'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(EQ(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

LE(s(n), s(m)) -> LE(n, m)

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LE(s(n), s(m)) -> LE(n, m)

one new Dependency Pair is created:

LE(s(s(n'')), s(s(m''))) -> LE(s(n''), s(m''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

LE(s(s(n'')), s(s(m''))) -> LE(s(n''), s(m''))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LE(s(s(n'')), s(s(m''))) -> LE(s(n''), s(m''))

one new Dependency Pair is created:

LE(s(s(s(n''''))), s(s(s(m'''')))) -> LE(s(s(n'''')), s(s(m'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 10

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

LE(s(s(s(n''''))), s(s(s(m'''')))) -> LE(s(s(n'''')), s(s(m'''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(s(s(n''''))), s(s(s(m'''')))) -> LE(s(s(n'''')), s(s(m'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))

four new Dependency Pairs are created:

REPLACE(0, m, cons(0, x)) -> IF_REPLACE(true, 0, m, cons(0, x))
REPLACE(0, m, cons(s(m''), x)) -> IF_REPLACE(false, 0, m, cons(s(m''), x))
REPLACE(s(n''), m, cons(0, x)) -> IF_REPLACE(false, s(n''), m, cons(0, x))
REPLACE(s(n''), m, cons(s(m''), x)) -> IF_REPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Narrowing Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(n''), m, cons(s(m''), x)) -> IF_REPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))
REPLACE(s(n''), m, cons(0, x)) -> IF_REPLACE(false, s(n''), m, cons(0, x))
REPLACE(0, m, cons(s(m''), x)) -> IF_REPLACE(false, 0, m, cons(s(m''), x))
IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

REPLACE(s(n''), m, cons(s(m''), x)) -> IF_REPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))

four new Dependency Pairs are created:

REPLACE(s(0), m, cons(s(0), x)) -> IF_REPLACE(true, s(0), m, cons(s(0), x))
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 13

                 ↳Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
REPLACE(0, m, cons(s(m''), x)) -> IF_REPLACE(false, 0, m, cons(s(m''), x))
IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
REPLACE(s(n''), m, cons(0, x)) -> IF_REPLACE(false, s(n''), m, cons(0, x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)

five new Dependency Pairs are created:

IF_REPLACE(false, 0, m'', cons(s(m''''), x'')) -> REPLACE(0, m'', x'')
IF_REPLACE(false, s(n''''), m'', cons(0, x'')) -> REPLACE(s(n''''), m'', x'')
IF_REPLACE(false, s(0), m'', cons(s(s(m''''')), x'')) -> REPLACE(s(0), m'', x'')
IF_REPLACE(false, s(s(n''')), m'', cons(s(0), x'')) -> REPLACE(s(s(n''')), m'', x'')
IF_REPLACE(false, s(s(n''')), m'', cons(s(s(m''''')), x')) -> REPLACE(s(s(n''')), m'', x')

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n''')), m'', cons(s(0), x'')) -> REPLACE(s(s(n''')), m'', x'')
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(0), m'', cons(s(s(m''''')), x'')) -> REPLACE(s(0), m'', x'')
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(n''''), m'', cons(0, x'')) -> REPLACE(s(n''''), m'', x'')
REPLACE(s(n''), m, cons(0, x)) -> IF_REPLACE(false, s(n''), m, cons(0, x))
IF_REPLACE(false, s(s(n''')), m'', cons(s(s(m''''')), x')) -> REPLACE(s(s(n''')), m'', x')
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_REPLACE(false, s(n''''), m'', cons(0, x'')) -> REPLACE(s(n''''), m'', x'')

four new Dependency Pairs are created:

IF_REPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
IF_REPLACE(false, s(0), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(0), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(0), x'''))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(s(m''''')), x'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n''')), m'', cons(s(s(m''''')), x')) -> REPLACE(s(s(n''')), m'', x')
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(s(m''''')), x'''))
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(0), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(0), x'''))
IF_REPLACE(false, s(0), m'', cons(s(s(m''''')), x'')) -> REPLACE(s(0), m'', x'')
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(0), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n''), m, cons(0, x)) -> IF_REPLACE(false, s(n''), m, cons(0, x))
IF_REPLACE(false, s(s(n''')), m'', cons(s(0), x'')) -> REPLACE(s(s(n''')), m'', x'')

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

REPLACE(s(n''), m, cons(0, x)) -> IF_REPLACE(false, s(n''), m, cons(0, x))

four new Dependency Pairs are created:

REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IF_REPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
REPLACE(s(0), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(0), m', cons(0, cons(s(s(m''''''')), x''''')))
REPLACE(s(s(n''''')), m', cons(0, cons(s(0), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(0), x''''')))
REPLACE(s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(s(m''''')), x'''))
REPLACE(s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(0), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(0), x'''))
REPLACE(s(s(n''''')), m', cons(0, cons(s(0), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(0), x''''')))
IF_REPLACE(false, s(0), m'', cons(s(s(m''''')), x'')) -> REPLACE(s(0), m'', x'')
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(0), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(0), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IF_REPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(s(n''')), m'', cons(s(0), x'')) -> REPLACE(s(s(n''')), m'', x'')
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n''')), m'', cons(s(s(m''''')), x')) -> REPLACE(s(s(n''')), m'', x')

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_REPLACE(false, s(0), m'', cons(s(s(m''''')), x'')) -> REPLACE(s(0), m'', x'')

three new Dependency Pairs are created:

IF_REPLACE(false, s(0), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 20

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(0), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(0), x'''))
REPLACE(s(s(n''''')), m', cons(0, cons(s(0), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(0), x''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(0), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(0), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(0), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IF_REPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
IF_REPLACE(false, s(s(n''')), m'', cons(s(0), x'')) -> REPLACE(s(s(n''')), m'', x'')
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n''')), m'', cons(s(s(m''''')), x')) -> REPLACE(s(s(n''')), m'', x')
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(s(m''''')), x'''))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_REPLACE(false, s(s(n''')), m'', cons(s(0), x'')) -> REPLACE(s(s(n''')), m'', x'')

five new Dependency Pairs are created:

IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n''''')), m'''', cons(s(0), cons(0, cons(0, x''''''')))) -> REPLACE(s(s(n''''')), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(0), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(0), x''''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 21

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(0), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(0), x''''''')))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(0), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(0), x'''))
REPLACE(s(s(n''''')), m', cons(0, cons(s(0), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(0), x''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(0), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(0), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(0), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IF_REPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
IF_REPLACE(false, s(s(n''''')), m'''', cons(s(0), cons(0, cons(0, x''''''')))) -> REPLACE(s(s(n''''')), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n''')), m'', cons(s(s(m''''')), x')) -> REPLACE(s(s(n''')), m'', x')
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(s(m''''')), x'''))
REPLACE(s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x''''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_REPLACE(false, s(s(n''')), m'', cons(s(s(m''''')), x')) -> REPLACE(s(s(n''')), m'', x')

five new Dependency Pairs are created:

IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n''''')), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(s(n''''')), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(s(m''''')), cons(0, cons(s(0), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(0), x''''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 22

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n'''')), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(s(m''''')), cons(0, cons(s(0), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(0), x''''''')))
IF_REPLACE(false, s(s(n''''')), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(s(n''''')), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(0), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(0), x''''''')))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(0), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(0), x'''))
REPLACE(s(s(n''''')), m', cons(0, cons(s(0), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(0), x''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(0), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(0), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(0), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IF_REPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
IF_REPLACE(false, s(s(n''''')), m'''', cons(s(0), cons(0, cons(0, x''''''')))) -> REPLACE(s(s(n''''')), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(s(m''''')), x'''))
REPLACE(s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(0), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(0), x'''))
IF_REPLACE(false, s(0), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
IF_REPLACE(false, s(s(n''')), m''', cons(0, cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n''')), m''', cons(s(s(m''''')), x'''))

Additionally, the following usable rules for innermost can be oriented:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REPLACE(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 1

POL(false) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(IF_REPLACE(x₁, x₂, x₃, x₄)) = x₄

POL(true) = 0

POL(s(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 24

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n'''')), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(s(m''''')), cons(0, cons(s(0), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(0), x''''''')))
IF_REPLACE(false, s(s(n''''')), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(s(n''''')), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(0), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(0), x''''''')))
REPLACE(s(s(n''''')), m', cons(0, cons(s(0), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(0), x''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))
IF_REPLACE(false, s(0), m'''', cons(s(s(m''''')), cons(0, cons(0, x''''''')))) -> REPLACE(s(0), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(0), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))
REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
REPLACE(s(0), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(0), m', cons(0, cons(s(s(m''''''')), x''''')))
REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IF_REPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
IF_REPLACE(false, s(s(n''''')), m'''', cons(s(0), cons(0, cons(0, x''''''')))) -> REPLACE(s(s(n''''')), m'''', cons(0, cons(0, x''''''')))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
REPLACE(s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x'''''))) -> IF_REPLACE(false, s(s(n''''')), m', cons(0, cons(s(s(m''''''')), x''''')))
IF_REPLACE(false, s(s(n'''')), m'''', cons(s(0), cons(0, cons(s(s(m''''''''')), x''''''')))) -> REPLACE(s(s(n'''')), m'''', cons(0, cons(s(s(m''''''''')), x''''''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 2 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 25

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(0), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))

Additionally, the following usable rules for innermost can be oriented:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REPLACE(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 1

POL(false) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(IF_REPLACE(x₁, x₂, x₃, x₄)) = x₄

POL(true) = 0

POL(s(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 27

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(s(n')), m, cons(s(0), x)) -> IF_REPLACE(false, s(s(n')), m, cons(s(0), x))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(0), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(0), x'''))
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 29

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))
REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_REPLACE(false, s(s(n'''')), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(s(n'''')), m''', cons(s(s(m''''')), x'''))

Additionally, the following usable rules for innermost can be oriented:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REPLACE(x₁, x₂, x₃)) = x₃

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(false) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(IF_REPLACE(x₁, x₂, x₃, x₄)) = x₄

POL(true) = 0

POL(s(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 30

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

REPLACE(s(s(n')), m, cons(s(s(m''')), x)) -> IF_REPLACE(eq(n', m'''), s(s(n')), m, cons(s(s(m''')), x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 26

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))
IF_REPLACE(false, s(0), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_REPLACE(false, s(0), m''', cons(s(s(m''''')), cons(s(s(m''''')), x'''))) -> REPLACE(s(0), m''', cons(s(s(m''''')), x'''))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REPLACE(x₁, x₂, x₃)) = x₃

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(false) = 0

POL(IF_REPLACE(x₁, x₂, x₃, x₄)) = x₄

POL(s(x₁)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 28

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

REPLACE(s(0), m, cons(s(s(m''')), x)) -> IF_REPLACE(false, s(0), m, cons(s(s(m''')), x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, 0, m'', cons(s(m''''), x'')) -> REPLACE(0, m'', x'')
REPLACE(0, m, cons(s(m''), x)) -> IF_REPLACE(false, 0, m, cons(s(m''), x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_REPLACE(false, 0, m'', cons(s(m''''), x'')) -> REPLACE(0, m'', x'')

one new Dependency Pair is created:

IF_REPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_REPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))
REPLACE(0, m, cons(s(m''), x)) -> IF_REPLACE(false, 0, m, cons(s(m''), x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

REPLACE(0, m, cons(s(m''), x)) -> IF_REPLACE(false, 0, m, cons(s(m''), x))

one new Dependency Pair is created:

REPLACE(0, m', cons(s(m'''), cons(s(m'''''''), x'''''))) -> IF_REPLACE(false, 0, m', cons(s(m'''), cons(s(m'''''''), x''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 19

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pairs:

REPLACE(0, m', cons(s(m'''), cons(s(m'''''''), x'''''))) -> IF_REPLACE(false, 0, m', cons(s(m'''), cons(s(m'''''''), x''''')))
IF_REPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_REPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REPLACE(x₁, x₂, x₃)) = x₃

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(false) = 0

POL(IF_REPLACE(x₁, x₂, x₃, x₄)) = x₄

POL(s(x₁)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Nar

...

               →DP Problem 23

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

Dependency Pair:

REPLACE(0, m', cons(s(m'''), cons(s(m'''''''), x'''''))) -> IF_REPLACE(false, 0, m', cons(s(m'''), cons(s(m'''''''), x''''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Narrowing Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

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

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

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

three new Dependency Pairs are created:

MIN(cons(0, cons(m'', x))) -> IF_MIN(true, cons(0, cons(m'', x)))
MIN(cons(s(n''), cons(0, x))) -> IF_MIN(false, cons(s(n''), cons(0, x)))
MIN(cons(s(n''), cons(s(m''), x))) -> IF_MIN(le(n'', m''), cons(s(n''), cons(s(m''), x)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Narrowing Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

MIN(cons(s(n''), cons(s(m''), x))) -> IF_MIN(le(n'', m''), cons(s(n''), cons(s(m''), x)))
MIN(cons(s(n''), cons(0, x))) -> IF_MIN(false, cons(s(n''), cons(0, x)))
IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
MIN(cons(0, cons(m'', x))) -> IF_MIN(true, cons(0, cons(m'', x)))
IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MIN(cons(s(n''), cons(s(m''), x))) -> IF_MIN(le(n'', m''), cons(s(n''), cons(s(m''), x)))

three new Dependency Pairs are created:

MIN(cons(s(0), cons(s(m'''), x))) -> IF_MIN(true, cons(s(0), cons(s(m'''), x)))
MIN(cons(s(s(n')), cons(s(0), x))) -> IF_MIN(false, cons(s(s(n')), cons(s(0), x)))
MIN(cons(s(s(n')), cons(s(s(m')), x))) -> IF_MIN(le(n', m'), cons(s(s(n')), cons(s(s(m')), x)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 32

                 ↳Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

MIN(cons(s(s(n')), cons(s(s(m')), x))) -> IF_MIN(le(n', m'), cons(s(s(n')), cons(s(s(m')), x)))
MIN(cons(s(s(n')), cons(s(0), x))) -> IF_MIN(false, cons(s(s(n')), cons(s(0), x)))
MIN(cons(s(0), cons(s(m'''), x))) -> IF_MIN(true, cons(s(0), cons(s(m'''), x)))
IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
MIN(cons(0, cons(m'', x))) -> IF_MIN(true, cons(0, cons(m'', x)))
IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))
MIN(cons(s(n''), cons(0, x))) -> IF_MIN(false, cons(s(n''), cons(0, x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))

three new Dependency Pairs are created:

IF_MIN(true, cons(0, cons(m', x''))) -> MIN(cons(0, x''))
IF_MIN(true, cons(s(0), cons(s(m'''''), x''))) -> MIN(cons(s(0), x''))
IF_MIN(true, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(n''')), x'))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 33

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(true, cons(0, cons(m', x''))) -> MIN(cons(0, x''))
MIN(cons(0, cons(m'', x))) -> IF_MIN(true, cons(0, cons(m'', x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MIN(true, cons(0, cons(m', x''))) -> MIN(cons(0, x''))

one new Dependency Pair is created:

IF_MIN(true, cons(0, cons(m', cons(m'''', x''')))) -> MIN(cons(0, cons(m'''', x''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 35

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(true, cons(0, cons(m', cons(m'''', x''')))) -> MIN(cons(0, cons(m'''', x''')))
MIN(cons(0, cons(m'', x))) -> IF_MIN(true, cons(0, cons(m'', x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

MIN(cons(0, cons(m'', x))) -> IF_MIN(true, cons(0, cons(m'', x)))

one new Dependency Pair is created:

MIN(cons(0, cons(m'''', cons(m'''''', x''''')))) -> IF_MIN(true, cons(0, cons(m'''', cons(m'''''', x'''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 38

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pairs:

MIN(cons(0, cons(m'''', cons(m'''''', x''''')))) -> IF_MIN(true, cons(0, cons(m'''', cons(m'''''', x'''''))))
IF_MIN(true, cons(0, cons(m', cons(m'''', x''')))) -> MIN(cons(0, cons(m'''', x''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MIN(true, cons(0, cons(m', cons(m'''', x''')))) -> MIN(cons(0, cons(m'''', x''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(MIN(x₁)) = x₁

POL(true) = 0

POL(IF_MIN(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 42

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

Dependency Pair:

MIN(cons(0, cons(m'''', cons(m'''''', x''''')))) -> IF_MIN(true, cons(0, cons(m'''', cons(m'''''', x'''''))))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 34

                 ↳Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(true, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(n''')), x'))
MIN(cons(s(s(n')), cons(s(0), x))) -> IF_MIN(false, cons(s(s(n')), cons(s(0), x)))
IF_MIN(true, cons(s(0), cons(s(m'''''), x''))) -> MIN(cons(s(0), x''))
MIN(cons(s(0), cons(s(m'''), x))) -> IF_MIN(true, cons(s(0), cons(s(m'''), x)))
MIN(cons(s(n''), cons(0, x))) -> IF_MIN(false, cons(s(n''), cons(0, x)))
IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))
MIN(cons(s(s(n')), cons(s(s(m')), x))) -> IF_MIN(le(n', m'), cons(s(s(n')), cons(s(s(m')), x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))

three new Dependency Pairs are created:

IF_MIN(false, cons(s(n''''), cons(0, x''))) -> MIN(cons(0, x''))
IF_MIN(false, cons(s(s(n''')), cons(s(0), x''))) -> MIN(cons(s(0), x''))
IF_MIN(false, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(m''')), x'))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 36

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(true, cons(s(0), cons(s(m'''''), x''))) -> MIN(cons(s(0), x''))
MIN(cons(s(0), cons(s(m'''), x))) -> IF_MIN(true, cons(s(0), cons(s(m'''), x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MIN(true, cons(s(0), cons(s(m'''''), x''))) -> MIN(cons(s(0), x''))

one new Dependency Pair is created:

IF_MIN(true, cons(s(0), cons(s(m'''''), cons(s(m'''''), x''')))) -> MIN(cons(s(0), cons(s(m'''''), x''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 39

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(true, cons(s(0), cons(s(m'''''), cons(s(m'''''), x''')))) -> MIN(cons(s(0), cons(s(m'''''), x''')))
MIN(cons(s(0), cons(s(m'''), x))) -> IF_MIN(true, cons(s(0), cons(s(m'''), x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MIN(true, cons(s(0), cons(s(m'''''), cons(s(m'''''), x''')))) -> MIN(cons(s(0), cons(s(m'''''), x''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(MIN(x₁)) = x₁

POL(true) = 0

POL(IF_MIN(x₁, x₂)) = x₂

POL(s(x₁)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 43

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

Dependency Pair:

MIN(cons(s(0), cons(s(m'''), x))) -> IF_MIN(true, cons(s(0), cons(s(m'''), x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 37

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(false, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(m''')), x'))
MIN(cons(s(s(n')), cons(s(s(m')), x))) -> IF_MIN(le(n', m'), cons(s(s(n')), cons(s(s(m')), x)))
IF_MIN(true, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(n''')), x'))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MIN(true, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(n''')), x'))

one new Dependency Pair is created:

IF_MIN(true, cons(s(s(n'''')), cons(s(s(m''')), cons(s(s(m''')), x''')))) -> MIN(cons(s(s(n'''')), cons(s(s(m''')), x''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 40

                 ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(true, cons(s(s(n'''')), cons(s(s(m''')), cons(s(s(m''')), x''')))) -> MIN(cons(s(s(n'''')), cons(s(s(m''')), x''')))
MIN(cons(s(s(n')), cons(s(s(m')), x))) -> IF_MIN(le(n', m'), cons(s(s(n')), cons(s(s(m')), x)))
IF_MIN(false, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(m''')), x'))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MIN(false, cons(s(s(n''')), cons(s(s(m''')), x'))) -> MIN(cons(s(s(m''')), x'))

one new Dependency Pair is created:

IF_MIN(false, cons(s(s(n''')), cons(s(s(m'''')), cons(s(s(m''0)), x''')))) -> MIN(cons(s(s(m'''')), cons(s(s(m''0)), x''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 41

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pairs:

IF_MIN(false, cons(s(s(n''')), cons(s(s(m'''')), cons(s(s(m''0)), x''')))) -> MIN(cons(s(s(m'''')), cons(s(s(m''0)), x''')))
MIN(cons(s(s(n')), cons(s(s(m')), x))) -> IF_MIN(le(n', m'), cons(s(s(n')), cons(s(s(m')), x)))
IF_MIN(true, cons(s(s(n'''')), cons(s(s(m''')), cons(s(s(m''')), x''')))) -> MIN(cons(s(s(n'''')), cons(s(s(m''')), x''')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_MIN(false, cons(s(s(n''')), cons(s(s(m'''')), cons(s(s(m''0)), x''')))) -> MIN(cons(s(s(m'''')), cons(s(s(m''0)), x''')))
IF_MIN(true, cons(s(s(n'''')), cons(s(s(m''')), cons(s(s(m''')), x''')))) -> MIN(cons(s(s(n'''')), cons(s(s(m''')), x''')))

Additionally, the following usable rules for innermost can be oriented:

le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(false) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(MIN(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 0

POL(IF_MIN(x₁, x₂)) = x₂

POL(le(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 31

             ↳Nar

...

               →DP Problem 44

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

Dependency Pair:

MIN(cons(s(s(n')), cons(s(s(m')), x))) -> IF_MIN(le(n', m'), cons(s(s(n')), cons(s(s(m')), x)))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Narrowing Transformation

Dependency Pair:

SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))

five new Dependency Pairs are created:

SORT(cons(n'', nil)) -> SORT(nil)
SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(min(cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))
SORT(cons(0, nil)) -> SORT(replace(0, 0, nil))
SORT(cons(s(n''), nil)) -> SORT(replace(s(n''), s(n''), nil))
SORT(cons(n'', cons(m', x''))) -> SORT(replace(if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 45

             ↳Rewriting Transformation

Dependency Pairs:

SORT(cons(n'', cons(m', x''))) -> SORT(replace(if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))
SORT(cons(s(n''), nil)) -> SORT(replace(s(n''), s(n''), nil))
SORT(cons(0, nil)) -> SORT(replace(0, 0, nil))
SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(min(cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(min(cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))

one new Dependency Pair is created:

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 45

             ↳Rw

...

               →DP Problem 46

                 ↳Rewriting Transformation

Dependency Pairs:

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))
SORT(cons(s(n''), nil)) -> SORT(replace(s(n''), s(n''), nil))
SORT(cons(0, nil)) -> SORT(replace(0, 0, nil))
SORT(cons(n'', cons(m', x''))) -> SORT(replace(if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(0, nil)) -> SORT(replace(0, 0, nil))

one new Dependency Pair is created:

SORT(cons(0, nil)) -> SORT(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 45

             ↳Rw

...

               →DP Problem 47

                 ↳Rewriting Transformation

Dependency Pairs:

SORT(cons(n'', cons(m', x''))) -> SORT(replace(if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))
SORT(cons(s(n''), nil)) -> SORT(replace(s(n''), s(n''), nil))
SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(s(n''), nil)) -> SORT(replace(s(n''), s(n''), nil))

one new Dependency Pair is created:

SORT(cons(s(n''), nil)) -> SORT(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 45

             ↳Rw

...

               →DP Problem 48

                 ↳Rewriting Transformation

Dependency Pairs:

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))
SORT(cons(n'', cons(m', x''))) -> SORT(replace(if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(n'', cons(m', x''))) -> SORT(replace(if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))

one new Dependency Pair is created:

SORT(cons(n'', cons(m', x''))) -> SORT(if_replace(eq(if_min(le(n'', m'), cons(n'', cons(m', x''))), m'), if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 45

             ↳Rw

...

               →DP Problem 49

                 ↳Rewriting Transformation

Dependency Pairs:

SORT(cons(n'', cons(m', x''))) -> SORT(if_replace(eq(if_min(le(n'', m'), cons(n'', cons(m', x''))), m'), if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))
SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), min(cons(n'', cons(k', x''))), n'', cons(k', x'')))

one new Dependency Pair is created:

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), if_min(le(n'', k'), cons(n'', cons(k', x''))), n'', cons(k', x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 45

             ↳Rw

...

               →DP Problem 50

                 ↳Polynomial Ordering

Dependency Pairs:

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), if_min(le(n'', k'), cons(n'', cons(k', x''))), n'', cons(k', x'')))
SORT(cons(n'', cons(m', x''))) -> SORT(if_replace(eq(if_min(le(n'', m'), cons(n'', cons(m', x''))), m'), if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

SORT(cons(n'', cons(k', x''))) -> SORT(if_replace(eq(if_min(le(n'', k'), cons(n'', cons(k', x''))), k'), if_min(le(n'', k'), cons(n'', cons(k', x''))), n'', cons(k', x'')))
SORT(cons(n'', cons(m', x''))) -> SORT(if_replace(eq(if_min(le(n'', m'), cons(n'', cons(m', x''))), m'), if_min(le(n'', m'), cons(n'', cons(m', x''))), n'', cons(m', x'')))

Additionally, the following usable rules for innermost can be oriented:

if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(false) = 0

POL(true) = 0

POL(replace(x₁, x₂, x₃)) = x₃

POL(if_min(x₁, x₂)) = 0

POL(SORT(x₁)) = x₁

POL(eq(x₁, x₂)) = 0

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(min(x₁)) = 0

POL(nil) = 0

POL(s(x₁)) = 0

POL(le(x₁, x₂)) = 0

POL(if_replace(x₁, x₂, x₃, x₄)) = x₄

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

       →DP Problem 5

         ↳Nar

           →DP Problem 45

             ↳Rw

...

               →DP Problem 51

                 ↳Dependency Graph

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) -> cons(m, x)
if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:08 minutes

POL(REPLACE(x₁, x₂, x₃))	= x₃
POL(eq(x₁, x₂))	= 0
POL(0)	= 1
POL(false)	= 0
POL(cons(x₁, x₂))	= x₁ + x₂
POL(IF_REPLACE(x₁, x₂, x₃, x₄))	= x₄
POL(true)	= 0
POL(s(x₁))	= 0

POL(0)	= 0
POL(cons(x₁, x₂))	= 1 + x₂
POL(MIN(x₁))	= x₁
POL(true)	= 0
POL(IF_MIN(x₁, x₂))	= x₂