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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 MNOCProof (⇔)
    ↳14 QDP
    ↳15 DependencyGraphProof (⇔)
    ↳16 AND
      ↳17 QDP
        ↳18 UsableRulesProof (⇔)
        ↳19 QDP
        ↳20 QReductionProof (⇔)
        ↳21 QDP
        ↳22 QDPOrderProof (⇔)
        ↳23 QDP
        ↳24 PisEmptyProof (⇔)
        ↳25 TRUE
      ↳26 QDP
        ↳27 QDPSizeChangeProof (⇔)
        ↳28 TRUE
  ↳29 QDP
    ↳30 UsableRulesProof (⇔)
    ↳31 QDP
    ↳32 MNOCProof (⇔)
    ↳33 QDP
    ↳34 UsableRulesReductionPairsProof (⇔)
    ↳35 QDP
    ↳36 Rewriting (⇔)
    ↳37 QDP
    ↳38 UsableRulesProof (⇔)
    ↳39 QDP
    ↳40 QDPOrderProof (⇔)
    ↳41 QDP
    ↳42 PisEmptyProof (⇔)
    ↳43 TRUE
  ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 QDPOrderProof (⇔)
    ↳48 QDP
    ↳49 PisEmptyProof (⇔)
    ↳50 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(s(x), s(y)) → P(s(x))
MINUS(s(x), s(y)) → P(s(y))
MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) → MINUS(x, y)
P(s(s(x))) → P(s(x))
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(s(x), s(y)) → MINUS(x, y)
DIV(plus(x, y), z) → PLUS(div(x, z), div(y, z))
DIV(plus(x, y), z) → DIV(x, z)
DIV(plus(x, y), z) → DIV(y, z)
PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))
PLUS(s(x), y) → MINUS(s(x), s(0))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

P(s(s(x))) → P(s(x))

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

(8) 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:

  • P(s(s(x))) → P(s(x))
    The graph contains the following edges 1 > 1

(9) TRUE

(10) Obligation:

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

MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(x, plus(y, z)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(x, plus(y, z)) → MINUS(x, y)

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)

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

(13) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(14) Obligation:

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

MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(x, plus(y, z)) → MINUS(x, y)

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
minus(x0, plus(x1, x2))

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Complex Obligation (AND)

(17) Obligation:

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
minus(x0, plus(x1, x2))

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

(18) 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.

(19) Obligation:

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
minus(x0, plus(x1, x2))

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

(20) 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(0, x0)
minus(s(x0), s(x1))
minus(x0, plus(x1, x2))

(21) Obligation:

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

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

POL(s(x1)) =
/0\
\1/
 +
/01\
\01/
·x1

POL(p(x1)) =
/0\
\0/
 +
/11\
\10/
·x1

The following usable rules [FROCOS05] were oriented:

p(s(s(x))) → s(p(s(x)))

(23) Obligation:

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

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

MINUS(x, plus(y, z)) → MINUS(x, y)
MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
minus(x0, plus(x1, x2))

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

(27) 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(x, plus(y, z)) → MINUS(x, y)
    The graph contains the following edges 1 >= 1, 2 > 2

  • MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
    The graph contains the following edges 2 > 2

(28) TRUE

(29) Obligation:

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

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

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(30) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(31) Obligation:

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

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

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(x, 0) → x
minus(0, y) → 0
p(s(s(x))) → s(p(s(x)))

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

(32) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(33) Obligation:

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

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

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(x, 0) → x
minus(0, y) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

minus(s(x0), s(x1))
minus(x0, plus(x1, x2))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(34) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

minus(x, plus(y, z)) → minus(minus(x, y), z)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(PLUS(x1, x2)) = 2·x1 + 2·x2   
POL(minus(x1, x2)) = x1 + x2   
POL(p(x1)) = x1   
POL(plus(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(s(x1)) = x1   

(35) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(s(x0), s(x1))
minus(x0, plus(x1, x2))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(36) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule PLUS(s(x), y) → PLUS(y, minus(s(x), s(0))) at position [1] we obtained the following new rules [LPAR04]:

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

(37) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(s(x0), s(x1))
minus(x0, plus(x1, x2))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(38) 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.

(39) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(s(x0), s(x1))
minus(x0, plus(x1, x2))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), y) → PLUS(y, minus(p(s(x)), p(s(0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(PLUS(x1, x2)) =
/0\
\1/
 +
/01\
\00/
·x1 +
/01\
\01/
·x2

POL(s(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(minus(x1, x2)) =
/0\
\0/
 +
/11\
\10/
·x1 +
/11\
\10/
·x2

POL(p(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(0) =
/1\
\1/

The following usable rules [FROCOS05] were oriented:

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

(41) Obligation:

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

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

The set Q consists of the following terms:

minus(s(x0), s(x1))
minus(x0, plus(x1, x2))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

DIV(plus(x, y), z) → DIV(x, z)
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(plus(x, y), z) → DIV(y, z)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DIV(plus(x, y), z) → DIV(x, z)
DIV(plus(x, y), z) → DIV(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(p(x1)) = x1   
POL(plus(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
minus(0, y) → 0
minus(x, 0) → x

(46) Obligation:

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

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

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(p(x1)) = x1   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
minus(0, y) → 0
minus(x, 0) → x

(48) Obligation:

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE