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)  =  -1(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
s1 > -^11


The following usable rules [FROCOS05] were oriented: none

(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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(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)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(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)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(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.