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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

(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(x, s(y)) → +1(x, y)
*1(x, s(y)) → +1(*(x, y), x)
*1(x, s(y)) → *1(x, y)
GE(s(x), s(y)) → GE(x, y)
-1(s(x), s(y)) → -1(x, y)
FACT(x) → IFFACT(x, ge(x, s(s(0))))
FACT(x) → GE(x, s(s(0)))
IFFACT(x, true) → *1(x, fact(-(x, s(0))))
IFFACT(x, true) → FACT(-(x, s(0)))
IFFACT(x, true) → -1(x, s(0))

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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 4 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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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.


GE(s(x), s(y)) → GE(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GE(x1, x2)  =  x2
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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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(x, s(y)) → +1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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(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

(19) Obligation:

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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:

IFFACT(x, true) → FACT(-(x, s(0)))
FACT(x) → IFFACT(x, ge(x, s(s(0))))

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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