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 Instantiation (⇔)
        ↳27 QDP
      ↳28 Narrowing (⇔)
        ↳29 QDP
        ↳30 DependencyGraphProof (⇔)
        ↳31 QDP
        ↳32 Narrowing (⇔)
        ↳33 QDP
        ↳34 DependencyGraphProof (⇔)
        ↳35 QDP
        ↳36 QDPOrderProof (⇔)
        ↳37 QDP
        ↳38 DependencyGraphProof (⇔)
        ↳39 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, s(y)) → if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) → s(quot(minus(x, y), y))
if_quot(false, x, y) → 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, s(y)) → if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) → s(quot(minus(x, y), y))
if_quot(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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:

MINUS(s(x), s(y)) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
QUOT(x, s(y)) → IF_QUOT(le(s(y), x), x, s(y))
QUOT(x, s(y)) → LE(s(y), x)
IF_QUOT(true, x, y) → QUOT(minus(x, y), y)
IF_QUOT(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, s(y)) → if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) → s(quot(minus(x, y), y))
if_quot(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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 3 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, s(y)) → if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) → s(quot(minus(x, y), y))
if_quot(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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:

LE(s(x), s(y)) → LE(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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].

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, x0, x1)

(11) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

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:

  • LE(s(x), s(y)) → LE(x, y)
    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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, s(y)) → if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) → s(quot(minus(x, y), y))
if_quot(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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].

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, x0, x1)

(18) 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.

(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:

  • MINUS(s(x), s(y)) → MINUS(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:

IF_QUOT(true, x, y) → QUOT(minus(x, y), y)
QUOT(x, s(y)) → IF_QUOT(le(s(y), x), x, s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, s(y)) → if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) → s(quot(minus(x, y), y))
if_quot(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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:

IF_QUOT(true, x, y) → QUOT(minus(x, y), y)
QUOT(x, s(y)) → IF_QUOT(le(s(y), x), x, s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, 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].

quot(x0, s(x1))
if_quot(true, x0, x1)
if_quot(false, x0, x1)

(25) Obligation:

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

IF_QUOT(true, x, y) → QUOT(minus(x, y), y)
QUOT(x, s(y)) → IF_QUOT(le(s(y), x), x, s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) Instantiation (EQUIVALENT transformation)

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

IF_QUOT(true, z0, s(z1)) → QUOT(minus(z0, s(z1)), s(z1))

(27) Obligation:

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

QUOT(x, s(y)) → IF_QUOT(le(s(y), x), x, s(y))
IF_QUOT(true, z0, s(z1)) → QUOT(minus(z0, s(z1)), s(z1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(28) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOT(x, s(y)) → IF_QUOT(le(s(y), x), x, s(y)) at position [0] we obtained the following new rules [LPAR04]:

QUOT(0, s(x0)) → IF_QUOT(false, 0, s(x0))
QUOT(s(x1), s(x0)) → IF_QUOT(le(x0, x1), s(x1), s(x0))

(29) Obligation:

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

IF_QUOT(true, x, y) → QUOT(minus(x, y), y)
QUOT(0, s(x0)) → IF_QUOT(false, 0, s(x0))
QUOT(s(x1), s(x0)) → IF_QUOT(le(x0, x1), s(x1), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(30) DependencyGraphProof (EQUIVALENT transformation)

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

(31) Obligation:

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

QUOT(s(x1), s(x0)) → IF_QUOT(le(x0, x1), s(x1), s(x0))
IF_QUOT(true, x, y) → QUOT(minus(x, y), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(32) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF_QUOT(true, x, y) → QUOT(minus(x, y), y) at position [0] we obtained the following new rules [LPAR04]:

IF_QUOT(true, x0, 0) → QUOT(x0, 0)
IF_QUOT(true, s(x0), s(x1)) → QUOT(minus(x0, x1), s(x1))

(33) Obligation:

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

QUOT(s(x1), s(x0)) → IF_QUOT(le(x0, x1), s(x1), s(x0))
IF_QUOT(true, x0, 0) → QUOT(x0, 0)
IF_QUOT(true, s(x0), s(x1)) → QUOT(minus(x0, x1), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(34) DependencyGraphProof (EQUIVALENT transformation)

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

(35) Obligation:

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

IF_QUOT(true, s(x0), s(x1)) → QUOT(minus(x0, x1), s(x1))
QUOT(s(x1), s(x0)) → IF_QUOT(le(x0, x1), s(x1), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IF_QUOT(true, s(x0), s(x1)) → QUOT(minus(x0, x1), s(x1))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF_QUOT(x1, x2, x3)) = x2   
POL(QUOT(x1, x2)) = x1   
POL(false) = 0   
POL(le(x1, x2)) = 0   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

(37) Obligation:

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

QUOT(s(x1), s(x0)) → IF_QUOT(le(x0, x1), s(x1), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(38) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(39) TRUE