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:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

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:

IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
INTLIST(x) → IF_INTLIST(empty(x), x)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INTLIST(false, x) → TAIL(x)
IF_INTLIST(false, x) → INTLIST(tail(x))
IF2(false, x, y) → P(x)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF_INTLIST(false, x) → HEAD(x)
IF2(false, x, y) → INTLIST(int(p(x), p(y)))
P(s(s(x))) → P(s(x))
INT(x, y) → ZERO(y)
INT(x, y) → ZERO(x)
IF2(false, x, y) → P(y)
IF2(false, x, y) → INT(p(x), p(y))
INTLIST(x) → EMPTY(x)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

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:

IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
INTLIST(x) → IF_INTLIST(empty(x), x)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INTLIST(false, x) → TAIL(x)
IF_INTLIST(false, x) → INTLIST(tail(x))
IF2(false, x, y) → P(x)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF_INTLIST(false, x) → HEAD(x)
IF2(false, x, y) → INTLIST(int(p(x), p(y)))
P(s(s(x))) → P(s(x))
INT(x, y) → ZERO(y)
INT(x, y) → ZERO(x)
IF2(false, x, y) → P(y)
IF2(false, x, y) → INT(p(x), p(y))
INTLIST(x) → EMPTY(x)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 8 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

INTLIST(x) → IF_INTLIST(empty(x), x)
IF_INTLIST(false, x) → INTLIST(tail(x))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

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.


INTLIST(x) → IF_INTLIST(empty(x), x)
The remaining pairs can at least be oriented weakly.

IF_INTLIST(false, x) → INTLIST(tail(x))
Used ordering: Polynomial interpretation [25,35]:

POL(IF_INTLIST(x1, x2)) = (1/4)x_1 + (3/4)x_2   
POL(empty(x1)) = 3/4 + (4)x_1   
POL(cons(x1, x2)) = 1/2 + (4)x_2   
POL(tail(x1)) = (1/4)x_1   
POL(true) = 3/4   
POL(false) = 1   
POL(INTLIST(x1)) = 1/4 + (2)x_1   
POL(nil) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

empty(nil) → true
tail(nil) → nil
empty(cons(x, y)) → false
tail(cons(x, y)) → y



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
          ↳ QDP
          ↳ QDP

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

IF_INTLIST(false, x) → INTLIST(tail(x))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

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.


P(s(s(x))) → P(s(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = (2)x_1   
POL(s(x1)) = 1/4 + (2)x_1   
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
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

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

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

IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF2(false, x, y) → INT(p(x), p(y))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

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