Termination w.r.t. Q proof of /tmp/tmp

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 AAECC Innermost (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

(0) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The TRS R 2 is

f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The signature Sigma is {f}

(2) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

+¹(s(x), y) → +¹(x, y)
*¹(x, s(y)) → +¹(x, *(x, y))
*¹(x, s(y)) → *¹(x, y)
TWICE(s(x)) → TWICE(x)
-¹(s(x), s(y)) → -¹(x, y)
F(s(x)) → F(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))
F(s(x)) → -¹(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))
F(s(x)) → *¹(s(s(x)), s(s(x)))
F(s(x)) → +¹(*(s(x), s(s(x))), s(s(0)))
F(s(x)) → *¹(s(x), s(s(x)))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2) = x2
s(x1) = s(x1)
+(x1, x2) = +(x1, x2)
0 = 0
*(x1, x2) = *(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = -(x1, x2)
f(x1) = f

Recursive path order with status [RPO].
Quasi-Precedence:

[0, f] > *₂ > +₂ > s₁
[0, f] > -₂ > s₁
twice₁ > s₁

Status:

s₁: multiset
+₂: [2,1]
0: multiset
*₂: [1,2]
twice₁: [1]
-₂: [2,1]
f: multiset

The following usable rules [FROCOS05] were oriented:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

(9) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

TWICE(s(x)) → TWICE(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TWICE(s(x)) → TWICE(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TWICE(x1) = x1
s(x1) = s(x1)
+(x1, x2) = +(x1, x2)
0 = 0
*(x1, x2) = *(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = -(x1, x2)
f(x1) = f

Recursive path order with status [RPO].
Quasi-Precedence:

[0, f] > *₂ > +₂ > s₁
[0, f] > -₂ > s₁
twice₁ > s₁

Status:

s₁: multiset
+₂: [2,1]
0: multiset
*₂: [1,2]
twice₁: [1]
-₂: [2,1]
f: multiset

The following usable rules [FROCOS05] were oriented:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

(14) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(s(x), y) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x1, x2)
s(x1) = s(x1)
+(x1, x2) = +(x1, x2)
0 = 0
*(x1, x2) = *(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = -(x1, x2)
f(x1) = f

Recursive path order with status [RPO].
Quasi-Precedence:

+^1₂ > s₁
[0, f] > *₂ > +₂ > s₁
[0, f] > -₂ > s₁
twice₁ > s₁

Status:

+^1₂: [2,1]
s₁: multiset
+₂: [2,1]
0: multiset
*₂: [1,2]
twice₁: [1]
-₂: [2,1]
f: multiset

The following usable rules [FROCOS05] were oriented:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

(19) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

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

*¹(x, s(y)) → *¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

*¹(x, s(y)) → *¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*¹(x1, x2) = *¹(x1, x2)
s(x1) = s(x1)
+(x1, x2) = +(x1, x2)
0 = 0
*(x1, x2) = *(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = -(x1, x2)
f(x1) = f

Recursive path order with status [RPO].
Quasi-Precedence:

*^1₂ > s₁
[0, f] > *₂ > +₂ > s₁
[0, f] > -₂ > s₁
twice₁ > s₁

Status:

*^1₂: [1,2]
s₁: multiset
+₂: [2,1]
0: multiset
*₂: [1,2]
twice₁: [1]
-₂: [2,1]
f: multiset

The following usable rules [FROCOS05] were oriented:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

(24) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE

(27) Obligation:

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

F(s(x)) → F(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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