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

(0) Obligation:

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

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 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:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(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:

LT(s(x), s(y)) → LT(x, y)
PLUS(x, s(y)) → PLUS(x, y)
QUOT(x, s(y)) → HELP(x, s(y), 0)
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
HELP(x, s(y), c) → LT(c, x)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
IF(true, x, s(y), c) → PLUS(c, s(y))

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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.


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

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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.


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

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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