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)))

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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

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)))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

EQ(x₁, x₂) -> EQ(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

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)))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

LE(x₁, x₂) -> LE(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

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)))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

IF_REPLACE(x₁, x₂, x₃, x₄) -> x₄
cons(x₁, x₂) -> cons(x₁, x₂)
REPLACE(x₁, x₂, x₃) -> x₃

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳AFS

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)))

The following dependency pairs can be strictly oriented:

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)))

The following usable rules w.r.t. to the AFS can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{cons, false, le} > {MIN, IF_MIN} > true
0 > true
s > true

resulting in one new DP problem.
Used Argument Filtering System:

MIN(x₁) -> MIN(x₁)
IF_MIN(x₁, x₂) -> IF_MIN(x₁, x₂)
cons(x₁, x₂) -> cons(x₁, x₂)
le(x₁, x₂) -> le(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 5

         ↳AFS

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Argument Filtering and Ordering

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)))

The following dependency pair can be strictly oriented:

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

The following usable rules w.r.t. to the AFS can be oriented:

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))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

SORT(x₁) -> SORT(x₁)
cons(x₁, x₂) -> cons(x₂)
replace(x₁, x₂, x₃) -> x₃
if_replace(x₁, x₂, x₃, x₄) -> x₄

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

           →DP Problem 10

             ↳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)))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:36 minutes