Termination w.r.t. Q proof of /tmp/tmpwND3vO/thiemann32.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:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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:

IF3(false, x, y) → GEN(s(x))
IF2(x, y) → 10¹
IF2(x, y) → LE(y, 10)
GEN(x) → LE(x, 10)
IF3(true, x, y) → IF2(x, s(y))
GEN(x) → 10¹
LE(s(x), s(y)) → LE(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
IF3(true, x, y) → TIMES(x, y)
TABLE → GEN(s(0))
TIMES(s(x), y) → TIMES(x, y)
GEN(x) → IF1(le(x, 10), x)
IF1(true, x) → IF2(x, x)
IF2(x, y) → IF3(le(y, 10), x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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:

IF3(false, x, y) → GEN(s(x))
IF2(x, y) → 10¹
IF2(x, y) → LE(y, 10)
GEN(x) → LE(x, 10)
IF3(true, x, y) → IF2(x, s(y))
GEN(x) → 10¹
LE(s(x), s(y)) → LE(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
IF3(true, x, y) → TIMES(x, y)
TABLE → GEN(s(0))
TIMES(s(x), y) → TIMES(x, y)
GEN(x) → IF1(le(x, 10), x)
IF1(true, x) → IF2(x, x)
IF2(x, y) → IF3(le(y, 10), x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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.

PLUS(s(x), y) → PLUS(x, y)

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

POL(PLUS(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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.

TIMES(s(x), y) → TIMES(x, y)

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

POL(TIMES(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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.

LE(s(x), s(y)) → LE(x, y)

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

POL(s(x₁)) = 1 + x_1
POL(LE(x₁, x₂)) = x_2

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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

          ↳ QDP

          ↳ QDP

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

IF3(false, x, y) → GEN(s(x))
IF3(true, x, y) → IF2(x, s(y))
GEN(x) → IF1(le(x, 10), x)
IF1(true, x) → IF2(x, x)
IF2(x, y) → IF3(le(y, 10), x, y)

The TRS R consists of the following rules:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

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