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 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP

(0) Obligation:

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

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

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:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(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:

TIMES(x, y) → SUM(generate(x, y))
TIMES(x, y) → GENERATE(x, y)
GENERATE(x, y) → GEN(x, y, 0)
GEN(x, y, z) → IF(ge(z, x), x, y, z)
GEN(x, y, z) → GE(z, x)
IF(false, x, y, z) → GEN(x, y, s(z))
SUM(cons(0, xs)) → SUM(xs)
SUM(cons(s(x), xs)) → SUM(cons(x, xs))
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) 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:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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.


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

(9) Obligation:

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

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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:

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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.


SUM(cons(0, xs)) → SUM(xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  SUM(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
0  =  0

Lexicographic Path Order [LPO].
Precedence:
cons2 > SUM1
0 > SUM1

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

SUM(cons(s(x), xs)) → SUM(cons(x, xs))

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
s1 > cons2

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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