Termination w.r.t. Q proof of /tmp/tmpRPXStD/ex15.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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:

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

↳ QTRS

  ↳ DependencyPairsProof

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:

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

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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:

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

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [13] we can delete all non-usable rules [14] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GT(s(u), s(v)) → GT(u, v)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GT(s(u), s(v)) → GT(u, v)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

GT(s(u), s(v)) → GT(u, v)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MAX(cons(u, l)) → MAX(l)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MAX(cons(u, l)) → MAX(l)

From the DPs we obtained the following set of size-change graphs:

MAX(cons(u, l)) → MAX(l)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

From the DPs we obtained the following set of size-change graphs:

EQUAL(s(x), s(y)) → EQUAL(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

MEMBER(n, cons(m, l)) → MEMBER(n, l)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

MEMBER(n, cons(m, l)) → MEMBER(n, l)

From the DPs we obtained the following set of size-change graphs:

MEMBER(n, cons(m, l)) → MEMBER(n, l)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

sort(x0)
cond2(false, x0, x1)
cond2(true, x0, x1)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule ST(n, l) → COND1(member(n, l), n, l) at position [0] we obtained the following new rules:

ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
ST(x0, nil) → COND1(false, x0, nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                    ↳ NonInfProof

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

ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
ST(x0, nil) → COND1(false, x0, nil)
COND1(true, n, l) → ST(s(n), l)
COND2(false, n, l) → ST(s(n), l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule COND1(false, n, l) → COND2(gt(n, max(l)), n, l) at position [0] we obtained the following new rules:

COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil)
COND1(false, 0, y1) → COND2(false, 0, y1)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                    ↳ NonInfProof

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

COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil)
COND1(false, 0, y1) → COND2(false, 0, y1)
ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
COND1(true, n, l) → ST(s(n), l)
ST(x0, nil) → COND1(false, x0, nil)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND1(true, n, l) → ST(s(n), l) we obtained the following new rules:

COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                    ↳ NonInfProof

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

COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil)
COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
COND1(false, 0, y1) → COND2(false, 0, y1)
ST(x0, nil) → COND1(false, x0, nil)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND2(false, n, l) → ST(s(n), l) we obtained the following new rules:

COND2(false, 0, z0) → ST(s(0), z0)
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND2(false, z0, nil) → ST(s(z0), nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                    ↳ NonInfProof

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

COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil)
COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND1(false, 0, y1) → COND2(false, 0, y1)
ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
COND2(false, 0, z0) → ST(s(0), z0)
ST(x0, nil) → COND1(false, x0, nil)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, z0, nil) → ST(s(z0), nil)
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2)) we obtained the following new rules:

ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                    ↳ NonInfProof

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

COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil)
COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND1(false, 0, y1) → COND2(false, 0, y1)
ST(x0, nil) → COND1(false, x0, nil)
COND2(false, 0, z0) → ST(s(0), z0)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND2(false, z0, nil) → ST(s(z0), nil)
ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule ST(x0, nil) → COND1(false, x0, nil) we obtained the following new rules:

ST(s(0), nil) → COND1(false, s(0), nil)
ST(s(z0), nil) → COND1(false, s(z0), nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                    ↳ NonInfProof

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

COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil)
COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND1(false, 0, y1) → COND2(false, 0, y1)
ST(s(0), nil) → COND1(false, s(0), nil)
COND2(false, 0, z0) → ST(s(0), z0)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, z0, nil) → ST(s(z0), nil)
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
ST(s(z0), nil) → COND1(false, s(z0), nil)
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                    ↳ NonInfProof

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

COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil)
COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
ST(s(0), nil) → COND1(false, s(0), nil)
COND2(false, 0, z0) → ST(s(0), z0)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, z0, nil) → ST(s(z0), nil)
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
ST(s(z0), nil) → COND1(false, s(z0), nil)
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule COND1(false, y0, nil) → COND2(gt(y0, 0), y0, nil) at position [0] we obtained the following new rules:

COND1(false, 0, nil) → COND2(false, 0, nil)
COND1(false, s(x0), nil) → COND2(true, s(x0), nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                    ↳ NonInfProof

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

COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND1(false, 0, nil) → COND2(false, 0, nil)
COND2(false, 0, z0) → ST(s(0), z0)
ST(s(0), nil) → COND1(false, s(0), nil)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND2(false, z0, nil) → ST(s(z0), nil)
ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
COND1(false, s(x0), nil) → COND2(true, s(x0), nil)
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))
ST(s(z0), nil) → COND1(false, s(z0), nil)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Instantiation

                    ↳ NonInfProof

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

COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND2(false, 0, z0) → ST(s(0), z0)
COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1))
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND1(false, y0, cons(x0, x1)) → COND2(gt(y0, if(gt(x0, max(x1)), x0, max(x1))), y0, cons(x0, x1)) we obtained the following new rules:

COND1(false, s(z0), cons(z1, z2)) → COND2(gt(s(z0), if(gt(z1, max(z2)), z1, max(z2))), s(z0), cons(z1, z2))
COND1(false, s(0), cons(z0, z1)) → COND2(gt(s(0), if(gt(z0, max(z1)), z0, max(z1))), s(0), cons(z0, z1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Instantiation

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                    ↳ NonInfProof

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

COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND2(false, 0, z0) → ST(s(0), z0)
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND1(false, s(z0), cons(z1, z2)) → COND2(gt(s(z0), if(gt(z1, max(z2)), z1, max(z2))), s(z0), cons(z1, z2))
ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
COND1(false, s(0), cons(z0, z1)) → COND2(gt(s(0), if(gt(z0, max(z1)), z0, max(z1))), s(0), cons(z0, z1))
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Instantiation

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                    ↳ NonInfProof

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

COND1(true, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND2(false, z0, cons(z1, z2)) → ST(s(z0), cons(z1, z2))
COND1(false, s(z0), cons(z1, z2)) → COND2(gt(s(z0), if(gt(z1, max(z2)), z1, max(z2))), s(z0), cons(z1, z2))
ST(s(z0), cons(z1, z2)) → COND1(or(equal(s(z0), z1), member(s(z0), z2)), s(z0), cons(z1, z2))
COND1(false, s(0), cons(z0, z1)) → COND2(gt(s(0), if(gt(z0, max(z1)), z0, max(z1))), s(0), cons(z0, z1))
ST(s(0), cons(x1, x2)) → COND1(or(equal(s(0), x1), member(s(0), x2)), s(0), cons(x1, x2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus with the following steps:
Note that final constraints are written in bold face.

For Pair ST(n, l) → COND1(member(n, l), n, l) the following chains were created:

We consider the chain ST(n, l) → COND1(member(n, l), n, l), COND1(true, n, l) → ST(s(n), l) which results in the following constraint:
- (1) (COND1(member(x2, x3), x2, x3)=COND1(true, x4, x5) ⇒ ST(x2, x3)≥COND1(member(x2, x3), x2, x3))
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
- (2) (member(x2, x3)=true ⇒ ST(x2, x3)≥COND1(member(x2, x3), x2, x3))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on member(x2, x3)=true which results in the following new constraint:
- (3) (or(equal(x43, x44), member(x43, x45))=true∧(member(x43, x45)=true ⇒ ST(x43, x45)≥COND1(member(x43, x45), x43, x45)) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
We simplified constraint (3) using rule (VII) which results in the following new constraint:
- (4) (equal(x43, x44)=x46∧member(x43, x45)=x47∧or(x46, x47)=true∧(member(x43, x45)=true ⇒ ST(x43, x45)≥COND1(member(x43, x45), x43, x45)) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on or(x46, x47)=true which results in the following new constraints:
- (5) (true=true∧equal(x43, x44)=x48∧member(x43, x45)=true∧(member(x43, x45)=true ⇒ ST(x43, x45)≥COND1(member(x43, x45), x43, x45)) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
- (6) (x49=true∧equal(x43, x44)=x49∧member(x43, x45)=false∧(member(x43, x45)=true ⇒ ST(x43, x45)≥COND1(member(x43, x45), x43, x45)) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
- (7) (equal(x43, x44)=x48∧member(x43, x45)=true∧(member(x43, x45)=true ⇒ ST(x43, x45)≥COND1(member(x43, x45), x43, x45)) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
- (8) (equal(x43, x44)=true ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (member(x43, x45)=true ⇒ ST(x43, x45)≥COND1(member(x43, x45), x43, x45)) with σ = [ ] which results in the following new constraint:
- (9) (equal(x43, x44)=x48∧ST(x43, x45)≥COND1(member(x43, x45), x43, x45) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
We simplified constraint (9) using rule (IV) which results in the following new constraint:
- (10) (ST(x43, x45)≥COND1(member(x43, x45), x43, x45) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on equal(x43, x44)=true which results in the following new constraints:
- (11) (true=true ⇒ ST(0, cons(0, x45))≥COND1(member(0, cons(0, x45)), 0, cons(0, x45)))
- (12) (equal(x52, x53)=true∧(∀x54:equal(x52, x53)=true ⇒ ST(x52, cons(x53, x54))≥COND1(member(x52, cons(x53, x54)), x52, cons(x53, x54))) ⇒ ST(s(x52), cons(s(x53), x45))≥COND1(member(s(x52), cons(s(x53), x45)), s(x52), cons(s(x53), x45)))
We simplified constraint (11) using rules (I), (II) which results in the following new constraint:
- (13) (ST(0, cons(0, x45))≥COND1(member(0, cons(0, x45)), 0, cons(0, x45)))
We simplified constraint (12) using rule (VI) where we applied the induction hypothesis (∀x54:equal(x52, x53)=true ⇒ ST(x52, cons(x53, x54))≥COND1(member(x52, cons(x53, x54)), x52, cons(x53, x54))) with σ = [x54 / x45] which results in the following new constraint:
- (14) (ST(x52, cons(x53, x45))≥COND1(member(x52, cons(x53, x45)), x52, cons(x53, x45)) ⇒ ST(s(x52), cons(s(x53), x45))≥COND1(member(s(x52), cons(s(x53), x45)), s(x52), cons(s(x53), x45)))
We consider the chain ST(n, l) → COND1(member(n, l), n, l), COND1(false, n, l) → COND2(gt(n, max(l)), n, l) which results in the following constraint:
- (15) (COND1(member(x8, x9), x8, x9)=COND1(false, x10, x11) ⇒ ST(x8, x9)≥COND1(member(x8, x9), x8, x9))
We simplified constraint (15) using rules (I), (II), (IV) which results in the following new constraint:
- (16) (member(x8, x9)=false ⇒ ST(x8, x9)≥COND1(member(x8, x9), x8, x9))
We simplified constraint (16) using rule (V) (with possible (I) afterwards) using induction on member(x8, x9)=false which results in the following new constraints:
- (17) (false=false ⇒ ST(x55, nil)≥COND1(member(x55, nil), x55, nil))
- (18) (or(equal(x56, x57), member(x56, x58))=false∧(member(x56, x58)=false ⇒ ST(x56, x58)≥COND1(member(x56, x58), x56, x58)) ⇒ ST(x56, cons(x57, x58))≥COND1(member(x56, cons(x57, x58)), x56, cons(x57, x58)))
We simplified constraint (17) using rules (I), (II) which results in the following new constraint:
- (19) (ST(x55, nil)≥COND1(member(x55, nil), x55, nil))
We simplified constraint (18) using rule (VII) which results in the following new constraint:
- (20) (equal(x56, x57)=x59∧member(x56, x58)=x60∧or(x59, x60)=false∧(member(x56, x58)=false ⇒ ST(x56, x58)≥COND1(member(x56, x58), x56, x58)) ⇒ ST(x56, cons(x57, x58))≥COND1(member(x56, cons(x57, x58)), x56, cons(x57, x58)))
We simplified constraint (20) using rule (V) (with possible (I) afterwards) using induction on or(x59, x60)=false which results in the following new constraint:
- (21) (x62=false∧equal(x56, x57)=x62∧member(x56, x58)=false∧(member(x56, x58)=false ⇒ ST(x56, x58)≥COND1(member(x56, x58), x56, x58)) ⇒ ST(x56, cons(x57, x58))≥COND1(member(x56, cons(x57, x58)), x56, cons(x57, x58)))
We simplified constraint (21) using rule (VI) where we applied the induction hypothesis (member(x56, x58)=false ⇒ ST(x56, x58)≥COND1(member(x56, x58), x56, x58)) with σ = [ ] which results in the following new constraint:
- (22) (x62=false∧equal(x56, x57)=x62∧ST(x56, x58)≥COND1(member(x56, x58), x56, x58) ⇒ ST(x56, cons(x57, x58))≥COND1(member(x56, cons(x57, x58)), x56, cons(x57, x58)))
We simplified constraint (22) using rule (III) which results in the following new constraint:
- (23) (equal(x56, x57)=false∧ST(x56, x58)≥COND1(member(x56, x58), x56, x58) ⇒ ST(x56, cons(x57, x58))≥COND1(member(x56, cons(x57, x58)), x56, cons(x57, x58)))
We simplified constraint (23) using rule (V) (with possible (I) afterwards) using induction on equal(x56, x57)=false which results in the following new constraints:
- (24) (false=false∧ST(s(x63), x58)≥COND1(member(s(x63), x58), s(x63), x58) ⇒ ST(s(x63), cons(0, x58))≥COND1(member(s(x63), cons(0, x58)), s(x63), cons(0, x58)))
- (25) (false=false∧ST(0, x58)≥COND1(member(0, x58), 0, x58) ⇒ ST(0, cons(s(x64), x58))≥COND1(member(0, cons(s(x64), x58)), 0, cons(s(x64), x58)))
- (26) (equal(x65, x66)=false∧ST(s(x65), x58)≥COND1(member(s(x65), x58), s(x65), x58)∧(∀x67:equal(x65, x66)=false∧ST(x65, x67)≥COND1(member(x65, x67), x65, x67) ⇒ ST(x65, cons(x66, x67))≥COND1(member(x65, cons(x66, x67)), x65, cons(x66, x67))) ⇒ ST(s(x65), cons(s(x66), x58))≥COND1(member(s(x65), cons(s(x66), x58)), s(x65), cons(s(x66), x58)))
We simplified constraint (24) using rules (I), (II) which results in the following new constraint:
- (27) (ST(s(x63), x58)≥COND1(member(s(x63), x58), s(x63), x58) ⇒ ST(s(x63), cons(0, x58))≥COND1(member(s(x63), cons(0, x58)), s(x63), cons(0, x58)))
We simplified constraint (25) using rules (I), (II) which results in the following new constraint:
- (28) (ST(0, x58)≥COND1(member(0, x58), 0, x58) ⇒ ST(0, cons(s(x64), x58))≥COND1(member(0, cons(s(x64), x58)), 0, cons(s(x64), x58)))
We simplified constraint (26) using rule (IV) which results in the following new constraint:
- (29) (equal(x65, x66)=false∧ST(s(x65), x58)≥COND1(member(s(x65), x58), s(x65), x58) ⇒ ST(s(x65), cons(s(x66), x58))≥COND1(member(s(x65), cons(s(x66), x58)), s(x65), cons(s(x66), x58)))
We simplified constraint (29) using rule (V) (with possible (I) afterwards) using induction on equal(x65, x66)=false which results in the following new constraints:
- (30) (false=false∧ST(s(s(x68)), x58)≥COND1(member(s(s(x68)), x58), s(s(x68)), x58) ⇒ ST(s(s(x68)), cons(s(0), x58))≥COND1(member(s(s(x68)), cons(s(0), x58)), s(s(x68)), cons(s(0), x58)))
- (31) (false=false∧ST(s(0), x58)≥COND1(member(s(0), x58), s(0), x58) ⇒ ST(s(0), cons(s(s(x69)), x58))≥COND1(member(s(0), cons(s(s(x69)), x58)), s(0), cons(s(s(x69)), x58)))
- (32) (equal(x70, x71)=false∧ST(s(s(x70)), x58)≥COND1(member(s(s(x70)), x58), s(s(x70)), x58)∧(∀x72:equal(x70, x71)=false∧ST(s(x70), x72)≥COND1(member(s(x70), x72), s(x70), x72) ⇒ ST(s(x70), cons(s(x71), x72))≥COND1(member(s(x70), cons(s(x71), x72)), s(x70), cons(s(x71), x72))) ⇒ ST(s(s(x70)), cons(s(s(x71)), x58))≥COND1(member(s(s(x70)), cons(s(s(x71)), x58)), s(s(x70)), cons(s(s(x71)), x58)))
We simplified constraint (30) using rules (I), (II) which results in the following new constraint:
- (33) (ST(s(s(x68)), x58)≥COND1(member(s(s(x68)), x58), s(s(x68)), x58) ⇒ ST(s(s(x68)), cons(s(0), x58))≥COND1(member(s(s(x68)), cons(s(0), x58)), s(s(x68)), cons(s(0), x58)))
We simplified constraint (31) using rules (I), (II) which results in the following new constraint:
- (34) (ST(s(0), x58)≥COND1(member(s(0), x58), s(0), x58) ⇒ ST(s(0), cons(s(s(x69)), x58))≥COND1(member(s(0), cons(s(s(x69)), x58)), s(0), cons(s(s(x69)), x58)))
We simplified constraint (32) using rule (IV) which results in the following new constraint:
- (35) (equal(x70, x71)=false∧ST(s(s(x70)), x58)≥COND1(member(s(s(x70)), x58), s(s(x70)), x58) ⇒ ST(s(s(x70)), cons(s(s(x71)), x58))≥COND1(member(s(s(x70)), cons(s(s(x71)), x58)), s(s(x70)), cons(s(s(x71)), x58)))
We simplified constraint (35) using rule (V) (with possible (I) afterwards) using induction on equal(x70, x71)=false which results in the following new constraints:
- (36) (false=false∧ST(s(s(s(x73))), x58)≥COND1(member(s(s(s(x73))), x58), s(s(s(x73))), x58) ⇒ ST(s(s(s(x73))), cons(s(s(0)), x58))≥COND1(member(s(s(s(x73))), cons(s(s(0)), x58)), s(s(s(x73))), cons(s(s(0)), x58)))
- (37) (false=false∧ST(s(s(0)), x58)≥COND1(member(s(s(0)), x58), s(s(0)), x58) ⇒ ST(s(s(0)), cons(s(s(s(x74))), x58))≥COND1(member(s(s(0)), cons(s(s(s(x74))), x58)), s(s(0)), cons(s(s(s(x74))), x58)))
- (38) (equal(x75, x76)=false∧ST(s(s(s(x75))), x58)≥COND1(member(s(s(s(x75))), x58), s(s(s(x75))), x58)∧(∀x77:equal(x75, x76)=false∧ST(s(s(x75)), x77)≥COND1(member(s(s(x75)), x77), s(s(x75)), x77) ⇒ ST(s(s(x75)), cons(s(s(x76)), x77))≥COND1(member(s(s(x75)), cons(s(s(x76)), x77)), s(s(x75)), cons(s(s(x76)), x77))) ⇒ ST(s(s(s(x75))), cons(s(s(s(x76))), x58))≥COND1(member(s(s(s(x75))), cons(s(s(s(x76))), x58)), s(s(s(x75))), cons(s(s(s(x76))), x58)))
We simplified constraint (36) using rules (I), (II) which results in the following new constraint:
- (39) (ST(s(s(s(x73))), x58)≥COND1(member(s(s(s(x73))), x58), s(s(s(x73))), x58) ⇒ ST(s(s(s(x73))), cons(s(s(0)), x58))≥COND1(member(s(s(s(x73))), cons(s(s(0)), x58)), s(s(s(x73))), cons(s(s(0)), x58)))
We simplified constraint (37) using rules (I), (II) which results in the following new constraint:
- (40) (ST(s(s(0)), x58)≥COND1(member(s(s(0)), x58), s(s(0)), x58) ⇒ ST(s(s(0)), cons(s(s(s(x74))), x58))≥COND1(member(s(s(0)), cons(s(s(s(x74))), x58)), s(s(0)), cons(s(s(s(x74))), x58)))
We simplified constraint (38) using rule (IV) which results in the following new constraint:
- (41) (ST(s(s(s(x75))), x58)≥COND1(member(s(s(s(x75))), x58), s(s(s(x75))), x58) ⇒ ST(s(s(s(x75))), cons(s(s(s(x76))), x58))≥COND1(member(s(s(s(x75))), cons(s(s(s(x76))), x58)), s(s(s(x75))), cons(s(s(s(x76))), x58)))

For Pair COND1(true, n, l) → ST(s(n), l) the following chains were created:

We consider the chain COND1(true, n, l) → ST(s(n), l), ST(n, l) → COND1(member(n, l), n, l) which results in the following constraint:
- (42) (ST(s(x12), x13)=ST(x14, x15) ⇒ COND1(true, x12, x13)≥ST(s(x12), x13))
We simplified constraint (42) using rules (I), (II), (IV) which results in the following new constraint:
- (43) (COND1(true, x12, x13)≥ST(s(x12), x13))

For Pair COND2(false, n, l) → ST(s(n), l) the following chains were created:

We consider the chain COND2(false, n, l) → ST(s(n), l), ST(n, l) → COND1(member(n, l), n, l) which results in the following constraint:
- (44) (ST(s(x22), x23)=ST(x24, x25) ⇒ COND2(false, x22, x23)≥ST(s(x22), x23))
We simplified constraint (44) using rules (I), (II), (IV) which results in the following new constraint:
- (45) (COND2(false, x22, x23)≥ST(s(x22), x23))

For Pair COND1(false, n, l) → COND2(gt(n, max(l)), n, l) the following chains were created:

We consider the chain COND1(false, n, l) → COND2(gt(n, max(l)), n, l), COND2(false, n, l) → ST(s(n), l) which results in the following constraint:
- (46) (COND2(gt(x36, max(x37)), x36, x37)=COND2(false, x38, x39) ⇒ COND1(false, x36, x37)≥COND2(gt(x36, max(x37)), x36, x37))
We simplified constraint (46) using rules (I), (II), (IV), (VII) which results in the following new constraint:
- (47) (max(x37)=x78∧gt(x36, x78)=false ⇒ COND1(false, x36, x37)≥COND2(gt(x36, max(x37)), x36, x37))
We simplified constraint (47) using rule (V) (with possible (I) afterwards) using induction on gt(x36, x78)=false which results in the following new constraints:
- (48) (false=false∧max(x37)=x79 ⇒ COND1(false, 0, x37)≥COND2(gt(0, max(x37)), 0, x37))
- (49) (gt(x81, x82)=false∧max(x37)=s(x82)∧(∀x83:gt(x81, x82)=false∧max(x83)=x82 ⇒ COND1(false, x81, x83)≥COND2(gt(x81, max(x83)), x81, x83)) ⇒ COND1(false, s(x81), x37)≥COND2(gt(s(x81), max(x37)), s(x81), x37))
We simplified constraint (48) using rules (I), (II), (IV) which results in the following new constraint:
- (50) (COND1(false, 0, x37)≥COND2(gt(0, max(x37)), 0, x37))
We simplified constraint (49) using rule (V) (with possible (I) afterwards) using induction on max(x37)=s(x82) which results in the following new constraint:
- (51) (if(gt(x84, max(x85)), x84, max(x85))=s(x82)∧gt(x81, x82)=false∧(∀x83:gt(x81, x82)=false∧max(x83)=x82 ⇒ COND1(false, x81, x83)≥COND2(gt(x81, max(x83)), x81, x83))∧(∀x86,x87,x88:max(x85)=s(x86)∧gt(x87, x86)=false∧(∀x88:gt(x87, x86)=false∧max(x88)=x86 ⇒ COND1(false, x87, x88)≥COND2(gt(x87, max(x88)), x87, x88)) ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) ⇒ COND1(false, s(x81), cons(x84, x85))≥COND2(gt(s(x81), max(cons(x84, x85))), s(x81), cons(x84, x85)))
We simplified constraint (51) using rules (IV), (VII) which results in the following new constraint:
- (52) (gt(x84, x91)=x89∧max(x85)=x90∧if(x89, x84, x90)=s(x82)∧gt(x81, x82)=false∧(∀x83:gt(x81, x82)=false∧max(x83)=x82 ⇒ COND1(false, x81, x83)≥COND2(gt(x81, max(x83)), x81, x83))∧(∀x86,x87,x88:max(x85)=s(x86)∧gt(x87, x86)=false∧(∀x88:gt(x87, x86)=false∧max(x88)=x86 ⇒ COND1(false, x87, x88)≥COND2(gt(x87, max(x88)), x87, x88)) ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) ⇒ COND1(false, s(x81), cons(x84, x85))≥COND2(gt(s(x81), max(cons(x84, x85))), s(x81), cons(x84, x85)))
We simplified constraint (52) using rule (V) (with possible (I) afterwards) using induction on if(x89, x84, x90)=s(x82) which results in the following new constraints:
- (53) (x92=s(x82)∧gt(x92, x91)=true∧max(x85)=x93∧gt(x81, x82)=false∧(∀x83:gt(x81, x82)=false∧max(x83)=x82 ⇒ COND1(false, x81, x83)≥COND2(gt(x81, max(x83)), x81, x83))∧(∀x86,x87,x88:max(x85)=s(x86)∧gt(x87, x86)=false∧(∀x88:gt(x87, x86)=false∧max(x88)=x86 ⇒ COND1(false, x87, x88)≥COND2(gt(x87, max(x88)), x87, x88)) ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) ⇒ COND1(false, s(x81), cons(x92, x85))≥COND2(gt(s(x81), max(cons(x92, x85))), s(x81), cons(x92, x85)))
- (54) (x95=s(x82)∧gt(x94, x91)=false∧max(x85)=x95∧gt(x81, x82)=false∧(∀x83:gt(x81, x82)=false∧max(x83)=x82 ⇒ COND1(false, x81, x83)≥COND2(gt(x81, max(x83)), x81, x83))∧(∀x86,x87,x88:max(x85)=s(x86)∧gt(x87, x86)=false∧(∀x88:gt(x87, x86)=false∧max(x88)=x86 ⇒ COND1(false, x87, x88)≥COND2(gt(x87, max(x88)), x87, x88)) ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) ⇒ COND1(false, s(x81), cons(x94, x85))≥COND2(gt(s(x81), max(cons(x94, x85))), s(x81), cons(x94, x85)))
We simplified constraint (53) using rules (III), (IV) which results in the following new constraint:
- (55) (gt(x81, x82)=false∧(∀x86,x87:max(x85)=s(x86)∧gt(x87, x86)=false ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) ⇒ COND1(false, s(x81), cons(s(x82), x85))≥COND2(gt(s(x81), max(cons(s(x82), x85))), s(x81), cons(s(x82), x85)))
We simplified constraint (54) using rule (III) which results in the following new constraint:
- (56) (gt(x94, x91)=false∧max(x85)=s(x82)∧gt(x81, x82)=false∧(∀x83:gt(x81, x82)=false∧max(x83)=x82 ⇒ COND1(false, x81, x83)≥COND2(gt(x81, max(x83)), x81, x83))∧(∀x86,x87,x88:max(x85)=s(x86)∧gt(x87, x86)=false∧(∀x88:gt(x87, x86)=false∧max(x88)=x86 ⇒ COND1(false, x87, x88)≥COND2(gt(x87, max(x88)), x87, x88)) ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) ⇒ COND1(false, s(x81), cons(x94, x85))≥COND2(gt(s(x81), max(cons(x94, x85))), s(x81), cons(x94, x85)))
We simplified constraint (55) using rule (V) (with possible (I) afterwards) using induction on gt(x81, x82)=false which results in the following new constraints:
- (57) (false=false∧(∀x86,x87:max(x85)=s(x86)∧gt(x87, x86)=false ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) ⇒ COND1(false, s(0), cons(s(x96), x85))≥COND2(gt(s(0), max(cons(s(x96), x85))), s(0), cons(s(x96), x85)))
- (58) (gt(x98, x99)=false∧(∀x86,x87:max(x85)=s(x86)∧gt(x87, x86)=false ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85))∧(∀x100,x101,x102:gt(x98, x99)=false∧(∀x101,x102:max(x100)=s(x101)∧gt(x102, x101)=false ⇒ COND1(false, s(x102), x100)≥COND2(gt(s(x102), max(x100)), s(x102), x100)) ⇒ COND1(false, s(x98), cons(s(x99), x100))≥COND2(gt(s(x98), max(cons(s(x99), x100))), s(x98), cons(s(x99), x100))) ⇒ COND1(false, s(s(x98)), cons(s(s(x99)), x85))≥COND2(gt(s(s(x98)), max(cons(s(s(x99)), x85))), s(s(x98)), cons(s(s(x99)), x85)))
We simplified constraint (57) using rules (I), (II), (IV) which results in the following new constraint:
- (59) (COND1(false, s(0), cons(s(x96), x85))≥COND2(gt(s(0), max(cons(s(x96), x85))), s(0), cons(s(x96), x85)))
We simplified constraint (58) using rule (VI) where we applied the induction hypothesis (∀x100,x101,x102:gt(x98, x99)=false∧(∀x101,x102:max(x100)=s(x101)∧gt(x102, x101)=false ⇒ COND1(false, s(x102), x100)≥COND2(gt(s(x102), max(x100)), s(x102), x100)) ⇒ COND1(false, s(x98), cons(s(x99), x100))≥COND2(gt(s(x98), max(cons(s(x99), x100))), s(x98), cons(s(x99), x100))) with σ = [x101 / x86, x102 / x87, x100 / x85] which results in the following new constraint:
- (60) (COND1(false, s(x98), cons(s(x99), x85))≥COND2(gt(s(x98), max(cons(s(x99), x85))), s(x98), cons(s(x99), x85)) ⇒ COND1(false, s(s(x98)), cons(s(s(x99)), x85))≥COND2(gt(s(s(x98)), max(cons(s(s(x99)), x85))), s(s(x98)), cons(s(s(x99)), x85)))
We simplified constraint (56) using rule (VI) where we applied the induction hypothesis (∀x86,x87,x88:max(x85)=s(x86)∧gt(x87, x86)=false∧(∀x88:gt(x87, x86)=false∧max(x88)=x86 ⇒ COND1(false, x87, x88)≥COND2(gt(x87, max(x88)), x87, x88)) ⇒ COND1(false, s(x87), x85)≥COND2(gt(s(x87), max(x85)), s(x87), x85)) with σ = [x88 / x83, x86 / x82, x87 / x81] which results in the following new constraint:
- (61) (gt(x94, x91)=false∧COND1(false, s(x81), x85)≥COND2(gt(s(x81), max(x85)), s(x81), x85) ⇒ COND1(false, s(x81), cons(x94, x85))≥COND2(gt(s(x81), max(cons(x94, x85))), s(x81), cons(x94, x85)))
We simplified constraint (61) using rule (V) (with possible (I) afterwards) using induction on gt(x94, x91)=false which results in the following new constraints:
- (62) (false=false∧COND1(false, s(x81), x85)≥COND2(gt(s(x81), max(x85)), s(x81), x85) ⇒ COND1(false, s(x81), cons(0, x85))≥COND2(gt(s(x81), max(cons(0, x85))), s(x81), cons(0, x85)))
- (63) (gt(x105, x106)=false∧COND1(false, s(x81), x85)≥COND2(gt(s(x81), max(x85)), s(x81), x85)∧(∀x107,x108:gt(x105, x106)=false∧COND1(false, s(x107), x108)≥COND2(gt(s(x107), max(x108)), s(x107), x108) ⇒ COND1(false, s(x107), cons(x105, x108))≥COND2(gt(s(x107), max(cons(x105, x108))), s(x107), cons(x105, x108))) ⇒ COND1(false, s(x81), cons(s(x105), x85))≥COND2(gt(s(x81), max(cons(s(x105), x85))), s(x81), cons(s(x105), x85)))
We simplified constraint (62) using rules (I), (II) which results in the following new constraint:
- (64) (COND1(false, s(x81), x85)≥COND2(gt(s(x81), max(x85)), s(x81), x85) ⇒ COND1(false, s(x81), cons(0, x85))≥COND2(gt(s(x81), max(cons(0, x85))), s(x81), cons(0, x85)))
We simplified constraint (63) using rule (VI) where we applied the induction hypothesis (∀x107,x108:gt(x105, x106)=false∧COND1(false, s(x107), x108)≥COND2(gt(s(x107), max(x108)), s(x107), x108) ⇒ COND1(false, s(x107), cons(x105, x108))≥COND2(gt(s(x107), max(cons(x105, x108))), s(x107), cons(x105, x108))) with σ = [x107 / x81, x108 / x85] which results in the following new constraint:
- (65) (COND1(false, s(x81), cons(x105, x85))≥COND2(gt(s(x81), max(cons(x105, x85))), s(x81), cons(x105, x85)) ⇒ COND1(false, s(x81), cons(s(x105), x85))≥COND2(gt(s(x81), max(cons(s(x105), x85))), s(x81), cons(s(x105), x85)))

To summarize, we get the following constraints P_≥ for the following pairs.

ST(n, l) → COND1(member(n, l), n, l)
- (ST(x43, x45)≥COND1(member(x43, x45), x43, x45) ⇒ ST(x43, cons(x44, x45))≥COND1(member(x43, cons(x44, x45)), x43, cons(x44, x45)))
- (ST(0, cons(0, x45))≥COND1(member(0, cons(0, x45)), 0, cons(0, x45)))
- (ST(x52, cons(x53, x45))≥COND1(member(x52, cons(x53, x45)), x52, cons(x53, x45)) ⇒ ST(s(x52), cons(s(x53), x45))≥COND1(member(s(x52), cons(s(x53), x45)), s(x52), cons(s(x53), x45)))
- (ST(x55, nil)≥COND1(member(x55, nil), x55, nil))
- (ST(s(x63), x58)≥COND1(member(s(x63), x58), s(x63), x58) ⇒ ST(s(x63), cons(0, x58))≥COND1(member(s(x63), cons(0, x58)), s(x63), cons(0, x58)))
- (ST(0, x58)≥COND1(member(0, x58), 0, x58) ⇒ ST(0, cons(s(x64), x58))≥COND1(member(0, cons(s(x64), x58)), 0, cons(s(x64), x58)))
- (ST(s(s(x68)), x58)≥COND1(member(s(s(x68)), x58), s(s(x68)), x58) ⇒ ST(s(s(x68)), cons(s(0), x58))≥COND1(member(s(s(x68)), cons(s(0), x58)), s(s(x68)), cons(s(0), x58)))
- (ST(s(0), x58)≥COND1(member(s(0), x58), s(0), x58) ⇒ ST(s(0), cons(s(s(x69)), x58))≥COND1(member(s(0), cons(s(s(x69)), x58)), s(0), cons(s(s(x69)), x58)))
- (ST(s(s(s(x73))), x58)≥COND1(member(s(s(s(x73))), x58), s(s(s(x73))), x58) ⇒ ST(s(s(s(x73))), cons(s(s(0)), x58))≥COND1(member(s(s(s(x73))), cons(s(s(0)), x58)), s(s(s(x73))), cons(s(s(0)), x58)))
- (ST(s(s(0)), x58)≥COND1(member(s(s(0)), x58), s(s(0)), x58) ⇒ ST(s(s(0)), cons(s(s(s(x74))), x58))≥COND1(member(s(s(0)), cons(s(s(s(x74))), x58)), s(s(0)), cons(s(s(s(x74))), x58)))
- (ST(s(s(s(x75))), x58)≥COND1(member(s(s(s(x75))), x58), s(s(s(x75))), x58) ⇒ ST(s(s(s(x75))), cons(s(s(s(x76))), x58))≥COND1(member(s(s(s(x75))), cons(s(s(s(x76))), x58)), s(s(s(x75))), cons(s(s(s(x76))), x58)))
COND1(true, n, l) → ST(s(n), l)
- (COND1(true, x12, x13)≥ST(s(x12), x13))
COND2(false, n, l) → ST(s(n), l)
- (COND2(false, x22, x23)≥ST(s(x22), x23))
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
- (COND1(false, 0, x37)≥COND2(gt(0, max(x37)), 0, x37))
- (COND1(false, s(x98), cons(s(x99), x85))≥COND2(gt(s(x98), max(cons(s(x99), x85))), s(x98), cons(s(x99), x85)) ⇒ COND1(false, s(s(x98)), cons(s(s(x99)), x85))≥COND2(gt(s(s(x98)), max(cons(s(s(x99)), x85))), s(s(x98)), cons(s(s(x99)), x85)))
- (COND1(false, s(x81), x85)≥COND2(gt(s(x81), max(x85)), s(x81), x85) ⇒ COND1(false, s(x81), cons(0, x85))≥COND2(gt(s(x81), max(cons(0, x85))), s(x81), cons(0, x85)))
- (COND1(false, s(x81), cons(x105, x85))≥COND2(gt(s(x81), max(cons(x105, x85))), s(x81), cons(x105, x85)) ⇒ COND1(false, s(x81), cons(s(x105), x85))≥COND2(gt(s(x81), max(cons(s(x105), x85))), s(x81), cons(s(x105), x85)))
- (COND1(false, s(0), cons(s(x96), x85))≥COND2(gt(s(0), max(cons(s(x96), x85))), s(0), cons(s(x96), x85)))

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved.
Polynomial interpretation [21]:

POL(0) = 0
POL(COND1(x₁, x₂, x₃)) = -1 - x₁ - x₂ + x₃
POL(COND2(x₁, x₂, x₃)) = -1 - x₁ - x₂ + x₃
POL(ST(x₁, x₂)) = -1 - x₁ + x₂
POL(c) = -1
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(equal(x₁, x₂)) = 0
POL(false) = 0
POL(gt(x₁, x₂)) = 0
POL(if(x₁, x₂, x₃)) = 1 + 2·x₁
POL(max(x₁)) = 2
POL(member(x₁, x₂)) = 0
POL(nil) = 0
POL(or(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

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

The following pairs are in P_bound:

COND1(false, n, l) → COND2(gt(n, max(l)), n, l)

The following rules are usable:

true → or(x, true)
false → gt(0, v)
false → member(n, nil)
gt(u, v) → gt(s(u), s(v))
x → or(x, false)
or(equal(n, m), member(n, l)) → member(n, cons(m, l))
true → gt(s(u), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                          ↳ DependencyGraphProof

                        ↳ QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

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

ST(n, l) → COND1(member(n, l), n, l)
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

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

ST(n, l) → COND1(member(n, l), n, l)
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

max(cons(x0, x1))
gt(s(x0), s(x1))
if(false, x0, x1)
max(nil)
gt(0, x0)
if(true, x0, x1)
gt(s(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ Narrowing

                                      ↳ NonInfProof

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

ST(n, l) → COND1(member(n, l), n, l)
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

equal(0, 0)
equal(s(x0), 0)
equal(s(x0), s(x1))
member(x0, cons(x1, x2))
member(x0, nil)
or(x0, false)
or(x0, true)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule ST(n, l) → COND1(member(n, l), n, l) at position [0] we obtained the following new rules:

ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
ST(x0, nil) → COND1(false, x0, nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                      ↳ NonInfProof

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

ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
ST(x0, nil) → COND1(false, x0, nil)
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

equal(0, 0)
equal(s(x0), 0)
equal(s(x0), s(x1))
member(x0, cons(x1, x2))
member(x0, nil)
or(x0, false)
or(x0, true)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Instantiation

                                      ↳ NonInfProof

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

ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2))
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

equal(0, 0)
equal(s(x0), 0)
equal(s(x0), s(x1))
member(x0, cons(x1, x2))
member(x0, nil)
or(x0, false)
or(x0, true)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule ST(x0, cons(x1, x2)) → COND1(or(equal(x0, x1), member(x0, x2)), x0, cons(x1, x2)) we obtained the following new rules:

ST(s(z0), cons(x1, x2)) → COND1(or(equal(s(z0), x1), member(s(z0), x2)), s(z0), cons(x1, x2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Instantiation

                                                ↳ QDP

                                      ↳ NonInfProof

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

COND1(true, n, l) → ST(s(n), l)
ST(s(z0), cons(x1, x2)) → COND1(or(equal(s(z0), x1), member(s(z0), x2)), s(z0), cons(x1, x2))

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

equal(0, 0)
equal(s(x0), 0)
equal(s(x0), s(x1))
member(x0, cons(x1, x2))
member(x0, nil)
or(x0, false)
or(x0, true)
equal(0, s(x0))

We consider the chain ST(n, l) → COND1(member(n, l), n, l), COND1(true, n, l) → ST(s(n), l) which results in the following constraint:
- (1) (COND1(member(x2, x3), x2, x3)=COND1(true, x4, x5) ⇒ ST(x2, x3)≥COND1(member(x2, x3), x2, x3))
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
- (2) (member(x2, x3)=true ⇒ ST(x2, x3)≥COND1(member(x2, x3), x2, x3))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on member(x2, x3)=true which results in the following new constraint:
- (3) (or(equal(x13, x14), member(x13, x15))=true∧(member(x13, x15)=true ⇒ ST(x13, x15)≥COND1(member(x13, x15), x13, x15)) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
We simplified constraint (3) using rule (VII) which results in the following new constraint:
- (4) (equal(x13, x14)=x16∧member(x13, x15)=x17∧or(x16, x17)=true∧(member(x13, x15)=true ⇒ ST(x13, x15)≥COND1(member(x13, x15), x13, x15)) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on or(x16, x17)=true which results in the following new constraints:
- (5) (true=true∧equal(x13, x14)=x18∧member(x13, x15)=true∧(member(x13, x15)=true ⇒ ST(x13, x15)≥COND1(member(x13, x15), x13, x15)) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
- (6) (x19=true∧equal(x13, x14)=x19∧member(x13, x15)=false∧(member(x13, x15)=true ⇒ ST(x13, x15)≥COND1(member(x13, x15), x13, x15)) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
- (7) (equal(x13, x14)=x18∧member(x13, x15)=true∧(member(x13, x15)=true ⇒ ST(x13, x15)≥COND1(member(x13, x15), x13, x15)) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
- (8) (equal(x13, x14)=true ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (member(x13, x15)=true ⇒ ST(x13, x15)≥COND1(member(x13, x15), x13, x15)) with σ = [ ] which results in the following new constraint:
- (9) (equal(x13, x14)=x18∧ST(x13, x15)≥COND1(member(x13, x15), x13, x15) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
We simplified constraint (9) using rule (IV) which results in the following new constraint:
- (10) (ST(x13, x15)≥COND1(member(x13, x15), x13, x15) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on equal(x13, x14)=true which results in the following new constraints:
- (11) (true=true ⇒ ST(0, cons(0, x15))≥COND1(member(0, cons(0, x15)), 0, cons(0, x15)))
- (12) (equal(x22, x23)=true∧(∀x24:equal(x22, x23)=true ⇒ ST(x22, cons(x23, x24))≥COND1(member(x22, cons(x23, x24)), x22, cons(x23, x24))) ⇒ ST(s(x22), cons(s(x23), x15))≥COND1(member(s(x22), cons(s(x23), x15)), s(x22), cons(s(x23), x15)))
We simplified constraint (11) using rules (I), (II) which results in the following new constraint:
- (13) (ST(0, cons(0, x15))≥COND1(member(0, cons(0, x15)), 0, cons(0, x15)))
We simplified constraint (12) using rule (VI) where we applied the induction hypothesis (∀x24:equal(x22, x23)=true ⇒ ST(x22, cons(x23, x24))≥COND1(member(x22, cons(x23, x24)), x22, cons(x23, x24))) with σ = [x24 / x15] which results in the following new constraint:
- (14) (ST(x22, cons(x23, x15))≥COND1(member(x22, cons(x23, x15)), x22, cons(x23, x15)) ⇒ ST(s(x22), cons(s(x23), x15))≥COND1(member(s(x22), cons(s(x23), x15)), s(x22), cons(s(x23), x15)))

For Pair COND1(true, n, l) → ST(s(n), l) the following chains were created:

We consider the chain COND1(true, n, l) → ST(s(n), l), ST(n, l) → COND1(member(n, l), n, l) which results in the following constraint:
- (15) (ST(s(x6), x7)=ST(x8, x9) ⇒ COND1(true, x6, x7)≥ST(s(x6), x7))
We simplified constraint (15) using rules (I), (II), (IV) which results in the following new constraint:
- (16) (COND1(true, x6, x7)≥ST(s(x6), x7))

To summarize, we get the following constraints P_≥ for the following pairs.

ST(n, l) → COND1(member(n, l), n, l)
- (ST(x13, x15)≥COND1(member(x13, x15), x13, x15) ⇒ ST(x13, cons(x14, x15))≥COND1(member(x13, cons(x14, x15)), x13, cons(x14, x15)))
- (ST(0, cons(0, x15))≥COND1(member(0, cons(0, x15)), 0, cons(0, x15)))
- (ST(x22, cons(x23, x15))≥COND1(member(x22, cons(x23, x15)), x22, cons(x23, x15)) ⇒ ST(s(x22), cons(s(x23), x15))≥COND1(member(s(x22), cons(s(x23), x15)), s(x22), cons(s(x23), x15)))
COND1(true, n, l) → ST(s(n), l)
- (COND1(true, x6, x7)≥ST(s(x6), x7))

POL(0) = 0
POL(COND1(x₁, x₂, x₃)) = -1 - x₁ - x₂ + x₃
POL(ST(x₁, x₂)) = -1 - x₁ + x₂
POL(c) = -1
POL(cons(x₁, x₂)) = x₁ + x₂
POL(equal(x₁, x₂)) = x₁
POL(false) = 0
POL(member(x₁, x₂)) = 0
POL(nil) = 2
POL(or(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

COND1(true, n, l) → ST(s(n), l)

The following pairs are in P_bound:

ST(n, l) → COND1(member(n, l), n, l)

The following rules are usable:

true → or(x, true)
or(equal(n, m), member(n, l)) → member(n, cons(m, l))
false → member(n, nil)
x → or(x, false)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ Narrowing

                                      ↳ NonInfProof

                                        ↳ AND

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                          ↳ QDP

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

ST(n, l) → COND1(member(n, l), n, l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

equal(0, 0)
equal(s(x0), 0)
equal(s(x0), s(x1))
member(x0, cons(x1, x2))
member(x0, nil)
or(x0, false)
or(x0, true)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ Narrowing

                                      ↳ NonInfProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

equal(0, 0)
equal(s(x0), 0)
equal(s(x0), s(x1))
member(x0, cons(x1, x2))
member(x0, nil)
or(x0, false)
or(x0, true)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.