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

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 UsableRulesProof (⇔)
↳6 QDP
↳7 QReductionProof (⇔)
↳8 QDP
↳9 QDPOrderProof (⇔)
↳10 QDP
↳11 QDPSizeChangeProof (⇔)
↳12 TRUE

(0) Obligation:

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

h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))

The signature Sigma is {h}

(2) Obligation:

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

h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))

The set Q consists of the following terms:

h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)

(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:

H(x, c(y, z)) → H(c(s(y), x), z)
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))

The TRS R consists of the following rules:

h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))

The set Q consists of the following terms:

h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)

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

(5) 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.

(6) Obligation:

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

H(x, c(y, z)) → H(c(s(y), x), z)
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))

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

h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)

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

(7) 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].

h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)

(8) Obligation:

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

H(x, c(y, z)) → H(c(s(y), x), z)
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

POL(c(x1, x2)) =
/0\
\0/
 +
/01\
\10/
·x1 +
/10\
\01/
·x2

POL(s(x1)) =
/0\
\0/
 +
/00\
\10/
·x1

POL(0) =
/1\
\0/

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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

H(x, c(y, z)) → H(c(s(y), x), z)

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

(11) 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:

  • H(x, c(y, z)) → H(c(s(y), x), z)
    The graph contains the following edges 2 > 2

(12) TRUE