Termination w.r.t. Q proof of /tmp/tmpqyBcHZ/thiemann07.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:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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:

IF2(true, l) → CAR(l)
IF2(false, l) → PROD(cons(times(car(l), cadr(l)), cddr(l)))
PLUS(x, y) → IFPLUS(isZero(x), x, y)
IF(false, b, l) → IF2(b, l)
IF2(false, l) → CAR(l)
TIMES(x, y) → ISZERO(x)
IFTIMES(false, x, y) → PLUS(y, times(p(x), y))
PLUS(x, y) → ISZERO(x)
TIMES(x, y) → IFTIMES(isZero(x), x, y)
PROD(l) → SHORTER(l, s(0))
IF2(false, l) → TIMES(car(l), cadr(l))
PROD(l) → SHORTER(l, 0)
IFPLUS(false, x, y) → PLUS(p(x), y)
SHORTER(cons(x, l), s(y)) → SHORTER(l, y)
PROD(l) → IF(shorter(l, 0), shorter(l, s(0)), l)
IF2(false, l) → CADR(l)
IFPLUS(false, x, y) → P(x)
IFTIMES(false, x, y) → P(x)
IFTIMES(false, x, y) → TIMES(p(x), y)
IF2(false, l) → CDDR(l)

The TRS R consists of the following rules:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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:

IF2(true, l) → CAR(l)
IF2(false, l) → PROD(cons(times(car(l), cadr(l)), cddr(l)))
PLUS(x, y) → IFPLUS(isZero(x), x, y)
IF(false, b, l) → IF2(b, l)
IF2(false, l) → CAR(l)
TIMES(x, y) → ISZERO(x)
IFTIMES(false, x, y) → PLUS(y, times(p(x), y))
PLUS(x, y) → ISZERO(x)
TIMES(x, y) → IFTIMES(isZero(x), x, y)
PROD(l) → SHORTER(l, s(0))
IF2(false, l) → TIMES(car(l), cadr(l))
PROD(l) → SHORTER(l, 0)
IFPLUS(false, x, y) → PLUS(p(x), y)
SHORTER(cons(x, l), s(y)) → SHORTER(l, y)
PROD(l) → IF(shorter(l, 0), shorter(l, s(0)), l)
IF2(false, l) → CADR(l)
IFPLUS(false, x, y) → P(x)
IFTIMES(false, x, y) → P(x)
IFTIMES(false, x, y) → TIMES(p(x), y)
IF2(false, l) → CDDR(l)

The TRS R consists of the following rules:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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 12 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SHORTER(cons(x, l), s(y)) → SHORTER(l, y)

The TRS R consists of the following rules:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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.

SHORTER(cons(x, l), s(y)) → SHORTER(l, y)

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

POL(cons(x₁, x₂)) = 1 + (4)x_2
POL(s(x₁)) = (4)x_1
POL(SHORTER(x₁, x₂)) = (4)x_1 + (3)x_2

The value of delta used in the strict ordering is 4.
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:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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:

PLUS(x, y) → IFPLUS(isZero(x), x, y)
IFPLUS(false, x, y) → PLUS(p(x), y)

The TRS R consists of the following rules:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TIMES(x, y) → IFTIMES(isZero(x), x, y)
IFTIMES(false, x, y) → TIMES(p(x), y)

The TRS R consists of the following rules:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF2(false, l) → PROD(cons(times(car(l), cadr(l)), cddr(l)))
IF(false, b, l) → IF2(b, l)
PROD(l) → IF(shorter(l, 0), shorter(l, s(0)), l)

The TRS R consists of the following rules:

car(cons(x, l)) → x
cddr(nil) → nil
cddr(cons(x, nil)) → nil
cddr(cons(x, cons(y, l))) → l
cadr(cons(x, cons(y, l))) → y
isZero(0) → true
isZero(s(x)) → false
plus(x, y) → ifplus(isZero(x), x, y)
ifplus(true, x, y) → y
ifplus(false, x, y) → s(plus(p(x), y))
times(x, y) → iftimes(isZero(x), x, y)
iftimes(true, x, y) → 0
iftimes(false, x, y) → plus(y, times(p(x), y))
p(s(x)) → x
p(0) → 0
shorter(nil, y) → true
shorter(cons(x, l), 0) → false
shorter(cons(x, l), s(y)) → shorter(l, y)
prod(l) → if(shorter(l, 0), shorter(l, s(0)), l)
if(true, b, l) → s(0)
if(false, b, l) → if2(b, l)
if2(true, l) → car(l)
if2(false, l) → prod(cons(times(car(l), cadr(l)), cddr(l)))

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