Termination w.r.t. Q proof of /tmp/tmpbVecc7/tpa08.trs

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

(0) Obligation:

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

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

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

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

The TRS R 2 is

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

The signature Sigma is {f}

(2) Obligation:

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

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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), s(y)) → -¹(x, y)
+¹(s(x), y) → +¹(x, y)
*¹(x, s(y)) → +¹(x, *(x, y))
*¹(x, s(y)) → *¹(x, y)
F(s(x)) → F(-(+(*(s(x), s(x)), *(s(x), s(s(s(0))))), *(s(s(x)), s(s(x)))))
F(s(x)) → -¹(+(*(s(x), s(x)), *(s(x), s(s(s(0))))), *(s(s(x)), s(s(x))))
F(s(x)) → +¹(*(s(x), s(x)), *(s(x), s(s(s(0)))))
F(s(x)) → *¹(s(x), s(x))
F(s(x)) → *¹(s(x), s(s(s(0))))
F(s(x)) → *¹(s(s(x)), s(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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 4 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) 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:

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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), 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)
s(x1) = s(x1)
-(x1, x2) = -(x1, x2)
0 = 0
+(x1, x2) = +(x1, x2)
*(x1, x2) = *(x1, x2)
f(x1) = f

Lexicographic path order with status [LPO].
Precedence:

+^1₁ > s₁
*₂ > 0 > s₁
*₂ > +₂ > s₁
f > -₂ > s₁

Status:

f: []
-₂: [2,1]
*₂: [1,2]
s₁: [1]
0: []
+₂: [1,2]
+^1₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

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

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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:

*¹(x, s(y)) → *¹(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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.

*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)
*(x1, x2) = *(x1, x2)
f(x1) = f

Lexicographic path order with status [LPO].
Precedence:

*₂ > +₂ > s₁
f > +₂ > s₁

Status:

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

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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), 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(x2)
s(x1) = s(x1)
-(x1, x2) = x1
0 = 0
+(x1, x2) = +(x1, x2)
*(x1, x2) = *(x1, x2)
f(x1) = f

Lexicographic path order with status [LPO].
Precedence:

*₂ > 0
*₂ > +₂ > s₁

Status:

f: []
*₂: [2,1]
-^1₁: [1]
s₁: [1]
0: []
+₂: [2,1]

The following usable rules [FROCOS05] were oriented:

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

(19) Obligation:

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

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(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:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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