Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔)
        ↳22 QDP
        ↳23 PisEmptyProof (⇔)
        ↳24 TRUE
    ↳25 QDP
        ↳26 QDPOrderProof (⇔)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔)
        ↳29 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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)

The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(n), s(m)) → LE(n, m)

The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


LE(s(n), s(m)) → LE(n, m)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
LE(x0, x1, x2)  =  LE(x0, x2)

Tags:
LE has argument tags [1,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
LE(x1, x2)  =  LE
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[LE, s1]

Status:
LE: []
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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))

The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MIN(x0, x1)  =  MIN(x1)
IF_MIN(x0, x1, x2)  =  IF_MIN(x0, x2)

Tags:
MIN has argument tags [7,0] and root tag 1
IF_MIN has argument tags [0,7,1] and root tag 0

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MIN(x1)  =  MIN
cons(x1, x2)  =  cons(x1, x2)
IF_MIN(x1, x2)  =  x2
le(x1, x2)  =  le(x1)
true  =  true
false  =  false
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
le1 > [MIN, false] > [cons2, true, 0]
s > [cons2, true, 0]

Status:
MIN: []
cons2: [2,1]
le1: [1]
true: []
false: []
0: []
s: []


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(n), s(m)) → EQ(n, m)

The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(s(n), s(m)) → EQ(n, m)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
EQ(x0, x1, x2)  =  EQ(x0, x2)

Tags:
EQ has argument tags [1,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
EQ(x1, x2)  =  EQ
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[EQ, s1]

Status:
EQ: []
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) TRUE

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
REPLACE(x0, x1, x2, x3)  =  REPLACE(x0, x1, x2)
IF_REPLACE(x0, x1, x2, x3, x4)  =  IF_REPLACE(x2, x3, x4)

Tags:
REPLACE has argument tags [7,12,0,11] and root tag 0
IF_REPLACE has argument tags [2,14,12,0,0] and root tag 0

Comparison: MS
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
REPLACE(x1, x2, x3)  =  x3
cons(x1, x2)  =  cons(x2)
IF_REPLACE(x1, x2, x3, x4)  =  x1
eq(x1, x2)  =  eq(x1, x2)
false  =  false
0  =  0
true  =  true
s(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
cons1 > [eq2, true] > false
0 > false

Status:
cons1: [1]
eq2: [1,2]
false: []
0: []
true: []


The following usable rules [FROCOS05] were oriented:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)

(22) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) TRUE

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SORT(cons(n, x)) → SORT(replace(min(cons(n, x)), n, x))

The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SORT(cons(n, x)) → SORT(replace(min(cons(n, x)), n, x))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
SORT(x0, x1)  =  SORT(x0)

Tags:
SORT has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
SORT(x1)  =  SORT(x1)
cons(x1, x2)  =  cons(x2)
replace(x1, x2, x3)  =  x3
min(x1)  =  min(x1)
0  =  0
nil  =  nil
s(x1)  =  s
if_min(x1, x2)  =  if_min(x1, x2)
le(x1, x2)  =  le(x1, x2)
true  =  true
false  =  false
if_replace(x1, x2, x3, x4)  =  x4
eq(x1, x2)  =  eq(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[SORT1, min1] > ifmin2 > [cons1, true] > 0 > s
[SORT1, min1] > ifmin2 > [cons1, true] > eq2 > s
[SORT1, min1] > le2 > false > [cons1, true] > 0 > s
[SORT1, min1] > le2 > false > [cons1, true] > eq2 > s
nil > 0 > s

Status:
SORT1: [1]
cons1: [1]
min1: [1]
0: []
nil: []
s: []
ifmin2: [1,2]
le2: [2,1]
true: []
false: []
eq2: [2,1]


The following usable rules [FROCOS05] were oriented:

replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)

(27) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(29) TRUE