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

(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))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) → x

The TRS R 2 is

f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x0))
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(-(p(*(s(x), s(x))), *(s(x), s(x))))
F(s(x)) → -¹(p(*(s(x), s(x))), *(s(x), s(x)))
F(s(x)) → P(*(s(x), s(x)))
F(s(x)) → *¹(s(x), 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))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x0))
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 4 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))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x0))
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)  =  x1
s(x1)  =  s(x1)
-(x1, x2)  =  -(x1, x2)
0  =  0
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  *(x1, x2)
p(x1)  =  x1
f(x1)  =  f

Lexicographic Path Order [LPO].
Precedence:

0 > [s₁, -₂, f]
*₂ > +₂ > [s₁, -₂, f]

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

Lexicographic Path Order [LPO].
Precedence:

0 > [s₁, -₂, f]
*₂ > +₂ > [s₁, -₂, f]

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

Lexicographic Path Order [LPO].
Precedence:

*₂ > 0
*₂ > +₂ > [-^1₁, s₁] > p₁
f > [-^1₁, s₁] > p₁

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))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x0))
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(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x0))
f(s(x0))

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