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

0 QTRS
↳1 AAECC Innermost (⇔)
↳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 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE

(0) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

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

The TRS R 2 is

f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The signature Sigma is {f}

(2) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(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:

-1(s(x), s(y)) → -1(x, y)
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) → P(-(s(x), y))
F(s(x), y) → -1(s(x), y)
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → -1(y, s(x))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) → P(-(x, s(y)))
F(x, s(y)) → -1(x, s(y))
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → -1(s(y), x)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(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 2 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(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:

-1(s(x), s(y)) → -1(x, y)

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

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(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].

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

(11) Obligation:

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

-1(s(x), s(y)) → -1(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:

  • -1(s(x), s(y)) → -1(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:

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(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:

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(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].

f(s(x0), x1)
f(x0, s(x1))

(18) Obligation:

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

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(F(x1, x2)) = (4)x2   
POL(s(x1)) = 4 + (4)x1   
POL(p(x1)) = (1/2)x1   
POL(-(x1, x2)) = x1   
POL(0) = 0   
The value of delta used in the strict ordering is 8.
The following usable rules [FROCOS05] were oriented:

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

(20) Obligation:

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

F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(F(x1, x2)) =
/1\
\0/
 +
/10\
\00/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/1\
\0/
 +
/11\
\11/
·x1

POL(p(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

POL(-(x1, x2)) =
/1\
\0/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(22) Obligation:

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

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

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE