Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
    ↳26 QDPSizeChangeProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 UsableRulesProof (⇔)
    ↳30 QDP
    ↳31 QReductionProof (⇔)
    ↳32 QDP
    ↳33 QDPSizeChangeProof (⇔)
    ↳34 TRUE
  ↳35 QDP
    ↳36 UsableRulesProof (⇔)
    ↳37 QDP
    ↳38 QReductionProof (⇔)
    ↳39 QDP
    ↳40 QDPSizeChangeProof (⇔)
    ↳41 TRUE
  ↳42 QDP
    ↳43 UsableRulesProof (⇔)
    ↳44 QDP
    ↳45 QReductionProof (⇔)
    ↳46 QDP
    ↳47 QDPSizeChangeProof (⇔)
    ↳48 TRUE
  ↳49 QDP
    ↳50 UsableRulesProof (⇔)
    ↳51 QDP
    ↳52 QReductionProof (⇔)
    ↳53 QDP
    ↳54 QDPSizeChangeProof (⇔)
    ↳55 TRUE
  ↳56 QDP
    ↳57 UsableRulesProof (⇔)
    ↳58 QDP
    ↳59 QReductionProof (⇔)
    ↳60 QDP
    ↳61 QDPSizeChangeProof (⇔)
    ↳62 TRUE
  ↳63 QDP
    ↳64 UsableRulesProof (⇔)
    ↳65 QDP
    ↳66 QReductionProof (⇔)
    ↳67 QDP
    ↳68 Narrowing (⇔)
    ↳69 QDP
    ↳70 DependencyGraphProof (⇔)
    ↳71 QDP
    ↳72 Narrowing (⇔)
    ↳73 QDP
    ↳74 DependencyGraphProof (⇔)
    ↳75 QDP
    ↳76 Instantiation (⇔)
    ↳77 QDP
    ↳78 Rewriting (⇔)
    ↳79 QDP
    ↳80 ForwardInstantiation (⇔)
    ↳81 QDP
    ↳82 Narrowing (⇔)
    ↳83 QDP
    ↳84 DependencyGraphProof (⇔)
    ↳85 QDP
    ↳86 Instantiation (⇔)
    ↳87 QDP
    ↳88 Rewriting (⇔)
    ↳89 QDP
    ↳90 Rewriting (⇔)
    ↳91 QDP
    ↳92 Instantiation (⇔)
    ↳93 QDP
    ↳94 DependencyGraphProof (⇔)
    ↳95 AND
      ↳96 QDP
        ↳97 ForwardInstantiation (⇔)
        ↳98 QDP
        ↳99 Rewriting (⇔)
        ↳100 QDP
        ↳101 QDPOrderProof (⇔)
        ↳102 QDP
        ↳103 DependencyGraphProof (⇔)
        ↳104 TRUE
      ↳105 QDP
        ↳106 UsableRulesProof (⇔)
        ↳107 QDP
        ↳108 ForwardInstantiation (⇔)
        ↳109 QDP
        ↳110 Rewriting (⇔)
        ↳111 QDP
        ↳112 MRRProof (⇔)
        ↳113 QDP
        ↳114 DependencyGraphProof (⇔)
        ↳115 TRUE

(0) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), 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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, 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:

GE(s(x), 0) → GE(x, 0)
GE(0, s(s(x))) → GE(0, s(x))
GE(s(x), s(y)) → GE(x, y)
MINUS(0, s(x)) → MINUS(0, x)
MINUS(s(x), 0) → MINUS(x, 0)
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(0, s(x)) → PLUS(0, x)
PLUS(s(x), y) → PLUS(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
DIV(x, y) → GE(y, s(0))
IFY(true, x, y) → IF(ge(x, y), x, y)
IFY(true, x, y) → GE(x, y)
IF(true, x, y) → DIV(minus(x, y), y)
IF(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, 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 9 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(0, s(x)) → PLUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

PLUS(0, s(x)) → PLUS(0, x)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(11) Obligation:

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

PLUS(0, s(x)) → PLUS(0, x)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • PLUS(0, s(x)) → PLUS(0, x)
    The graph contains the following edges 1 >= 1, 2 > 2

(13) TRUE

(14) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(18) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • PLUS(s(x), y) → PLUS(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(20) TRUE

(21) Obligation:

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

MINUS(s(x), 0) → MINUS(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

MINUS(s(x), 0) → MINUS(x, 0)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(25) Obligation:

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

MINUS(s(x), 0) → MINUS(x, 0)

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MINUS(s(x), 0) → MINUS(x, 0)
    The graph contains the following edges 1 > 1, 2 >= 2

(27) TRUE

(28) Obligation:

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

MINUS(0, s(x)) → MINUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

MINUS(0, s(x)) → MINUS(0, x)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(32) Obligation:

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

MINUS(0, s(x)) → MINUS(0, x)

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

(33) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MINUS(0, s(x)) → MINUS(0, x)
    The graph contains the following edges 1 >= 1, 2 > 2

(34) TRUE

(35) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(36) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(37) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(38) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(39) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

(40) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MINUS(s(x), s(y)) → MINUS(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(41) TRUE

(42) Obligation:

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

GE(0, s(s(x))) → GE(0, s(x))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(43) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(44) Obligation:

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

GE(0, s(s(x))) → GE(0, s(x))

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(45) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(46) Obligation:

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

GE(0, s(s(x))) → GE(0, s(x))

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

(47) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • GE(0, s(s(x))) → GE(0, s(x))
    The graph contains the following edges 1 >= 1, 2 > 2

(48) TRUE

(49) Obligation:

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

GE(s(x), 0) → GE(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(50) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(51) Obligation:

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

GE(s(x), 0) → GE(x, 0)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(52) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(53) Obligation:

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

GE(s(x), 0) → GE(x, 0)

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

(54) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • GE(s(x), 0) → GE(x, 0)
    The graph contains the following edges 1 > 1, 2 >= 2

(55) TRUE

(56) 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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(57) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(58) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(59) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(60) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

(61) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • GE(s(x), s(y)) → GE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(62) TRUE

(63) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(64) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(65) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(66) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(67) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(68) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule DIV(x, y) → IFY(ge(y, s(0)), x, y) at position [0] we obtained the following new rules [LPAR04]:

DIV(y0, 0) → IFY(false, y0, 0)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))

(69) Obligation:

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

IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, 0) → IFY(false, y0, 0)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(70) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(71) Obligation:

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(72) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IFY(true, x, y) → IF(ge(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, 0, 0) → IF(true, 0, 0)
IFY(true, s(x0), 0) → IF(ge(x0, 0), s(x0), 0)
IFY(true, 0, s(0)) → IF(false, 0, s(0))
IFY(true, 0, s(s(x0))) → IF(ge(0, s(x0)), 0, s(s(x0)))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))

(73) Obligation:

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, 0, 0) → IF(true, 0, 0)
IFY(true, s(x0), 0) → IF(ge(x0, 0), s(x0), 0)
IFY(true, 0, s(0)) → IF(false, 0, s(0))
IFY(true, 0, s(s(x0))) → IF(ge(0, s(x0)), 0, s(s(x0)))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(74) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(75) Obligation:

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(76) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, x, y) → DIV(minus(x, y), y) we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

(77) Obligation:

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(78) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

(79) Obligation:

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(80) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0)) we obtained the following new rules [LPAR04]:

DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))

(81) Obligation:

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(82) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(0), s(0)) → DIV(0, s(0))
IF(true, s(0), s(s(x0))) → DIV(minus(0, x0), s(s(x0)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

(83) Obligation:

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))
IF(true, s(0), s(0)) → DIV(0, s(0))
IF(true, s(0), s(s(x0))) → DIV(minus(0, x0), s(s(x0)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(84) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(85) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(86) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1)) we obtained the following new rules [LPAR04]:

DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))

(87) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(88) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0)) at position [0] we obtained the following new rules [LPAR04]:

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))

(89) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(90) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1))) at position [0] we obtained the following new rules [LPAR04]:

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))

(91) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(92) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1)) we obtained the following new rules [LPAR04]:

IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))

(93) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(94) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(95) Complex Obligation (AND)

(96) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(97) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1))) we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))

(98) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(99) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1))) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

(100) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(101) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(DIV(x1, x2)) =
/0\
\0/
 +
/01\
\11/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/0\
\0/
 +
/01\
\11/
·x1

POL(IFY(x1, x2, x3)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/01\
\11/
·x2 +
/00\
\00/
·x3

POL(ge(x1, x2)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/00\
\00/
·x2

POL(0) =
/0\
\1/

POL(IF(x1, x2, x3)) =
/0\
\0/
 +
/01\
\00/
·x1 +
/10\
\11/
·x2 +
/00\
\00/
·x3

POL(true) =
/0\
\1/

POL(minus(x1, x2)) =
/0\
\0/
 +
/01\
\11/
·x1 +
/00\
\00/
·x2

POL(false) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

ge(s(x), s(y)) → ge(x, y)
ge(0, s(s(x))) → ge(0, s(x))
ge(0, s(0)) → false
ge(s(x), 0) → ge(x, 0)
ge(0, 0) → true
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(minus(x, 0))
minus(0, s(x)) → minus(0, x)
minus(0, 0) → 0

(102) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(103) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(104) TRUE

(105) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(106) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(107) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(108) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0)) we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0))

(109) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(110) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0)) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

(111) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(112) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

Strictly oriented rules of the TRS R:

ge(s(x), 0) → ge(x, 0)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIV(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(IF(x1, x2, x3)) = 2 + x1 + x2 + 2·x3   
POL(IFY(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(ge(x1, x2)) = 2 + x1 + 2·x2   
POL(minus(x1, x2)) = x1 + x2   
POL(s(x1)) = 1 + 2·x1   
POL(true) = 2   

(113) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(114) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(115) TRUE