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 MNOCProof (⇔)
    ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 MNOCProof (⇔)
    ↳16 QDP
    ↳17 UsableRulesProof (⇔)
    ↳18 QDP
    ↳19 QReductionProof (⇔)
    ↳20 QDP
    ↳21 QDPSizeChangeProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 MNOCProof (⇔)
    ↳25 QDP
    ↳26 UsableRulesProof (⇔)
    ↳27 QDP
    ↳28 QReductionProof (⇔)
    ↳29 QDP
    ↳30 QDPSizeChangeProof (⇔)
    ↳31 TRUE
  ↳32 QDP
    ↳33 MNOCProof (⇔)
    ↳34 QDP
    ↳35 UsableRulesProof (⇔)
    ↳36 QDP
    ↳37 QReductionProof (⇔)
    ↳38 QDP
    ↳39 MRRProof (⇔)
    ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 PisEmptyProof (⇔)
    ↳44 TRUE

(0) Obligation:

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

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

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:

PLUS(x, s(y)) → PLUS(x, y)
TIMES(s(x), y) → PLUS(times(x, y), y)
TIMES(s(x), y) → TIMES(x, y)
P(s(s(x))) → P(s(x))
FAC(s(x)) → TIMES(fac(p(s(x))), s(x))
FAC(s(x)) → FAC(p(s(x)))
FAC(s(x)) → P(s(x))

The TRS R consists of the following rules:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

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 3 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:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

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

(6) MNOCProof (EQUIVALENT transformation)

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

(7) 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:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

The set Q consists of the following terms:

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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:

P(s(s(x))) → P(s(x))

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

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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].

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

(11) 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.

(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:

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

(13) TRUE

(14) 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:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

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

(15) MNOCProof (EQUIVALENT transformation)

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

(16) 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:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

The set Q consists of the following terms:

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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

(17) 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.

(18) Obligation:

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

PLUS(x, s(y)) → PLUS(x, y)

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

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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

(19) 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(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

(20) Obligation:

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

PLUS(x, s(y)) → PLUS(x, y)

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

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

(22) TRUE

(23) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

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

(24) MNOCProof (EQUIVALENT transformation)

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

(25) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

The set Q consists of the following terms:

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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

(26) 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.

(27) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

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

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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

(28) 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(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

(29) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

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

(30) 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:

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

(31) TRUE

(32) Obligation:

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

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

(33) MNOCProof (EQUIVALENT transformation)

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

(34) Obligation:

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

The set Q consists of the following terms:

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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

(35) 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.

(36) Obligation:

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))

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

(37) 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(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
fac(s(x0))

(38) Obligation:

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

p(s(s(x0)))
p(s(0))

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

(39) 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 rules of the TRS R:

p(s(0)) → 0

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(FAC(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   

(40) Obligation:

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

FAC(s(x)) → FAC(p(s(x)))

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)))
p(s(0))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(FAC(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(s(x1)) =
/1\
\0/
 +
/10\
\10/
·x1

POL(p(x1)) =
/0\
\1/
 +
/01\
\01/
·x1

The following usable rules [FROCOS05] were oriented:

p(s(s(x))) → s(p(s(x)))

(42) 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)))
p(s(0))

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

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) TRUE