Termination w.r.t. Q of the given QTRS could not be shown:

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

(0) Obligation:

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

SORT(l) → ST(0, l)
ST(n, l) → COND1(member(n, l), n, l)
ST(n, l) → MEMBER(n, l)
COND1(true, n, l) → ST(s(n), l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
COND1(false, n, l) → GT(n, max(l))
COND1(false, n, l) → MAX(l)
COND2(false, n, l) → ST(s(n), l)
MEMBER(n, cons(m, l)) → OR(equal(n, m), member(n, l))
MEMBER(n, cons(m, l)) → EQUAL(n, m)
MEMBER(n, cons(m, l)) → MEMBER(n, l)
EQUAL(s(x), s(y)) → EQUAL(x, y)
GT(s(u), s(v)) → GT(u, v)
MAX(cons(u, l)) → IF(gt(u, max(l)), u, max(l))
MAX(cons(u, l)) → GT(u, max(l))
MAX(cons(u, l)) → MAX(l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


GT(s(u), s(v)) → GT(u, v)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GT(x1, x2)  =  GT(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
s1 > GT1

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

MAX(cons(u, l)) → MAX(l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MAX(cons(u, l)) → MAX(l)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
cons2 > MAX1

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQUAL(s(x), s(y)) → EQUAL(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQUAL(x1, x2)  =  EQUAL(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
s1 > EQUAL1

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

MEMBER(n, cons(m, l)) → MEMBER(n, l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MEMBER(n, cons(m, l)) → MEMBER(n, l)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MEMBER(x1, x2)  =  MEMBER(x2)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
cons2 > MEMBER1

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

COND1(true, n, l) → ST(s(n), l)
ST(n, l) → COND1(member(n, l), n, l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
COND2(false, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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