Termination w.r.t. Q proof of /tmp/tmp98Gww9/24.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(true, x, y) → GR(x, 0)
COND4(false, x, y) → GR(x, 0)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND4(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(x, 0)
COND3(false, x, y) → GR(x, 0)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → GR(y, 0)
COND4(false, x, y) → GR(y, 0)
COND3(true, x, y) → P(x)
COND4(true, x, y) → P(y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND1(true, x, y) → GR(x, y)
COND4(false, x, y) → AND(gr(x, 0), gr(y, 0))

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.
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:

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(true, x, y) → GR(x, 0)
COND4(false, x, y) → GR(x, 0)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND4(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(x, 0)
COND3(false, x, y) → GR(x, 0)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → GR(y, 0)
COND4(false, x, y) → GR(y, 0)
COND3(true, x, y) → P(x)
COND4(true, x, y) → P(y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND1(true, x, y) → GR(x, y)
COND4(false, x, y) → AND(gr(x, 0), gr(y, 0))

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 13 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

GR(s(x), s(y)) → GR(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(GR(x₁, x₂)) = (3)x_2
POL(s(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

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