Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ AAECC Innermost

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.

We have applied [19,8] 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}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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)


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof

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

H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
H(x, c(y, z)) → H(c(s(y), 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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
H(x, c(y, z)) → H(c(s(y), 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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPOrderProof

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

H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
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.
We use the reduction pair processor [15].


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.

H(x, c(y, z)) → H(c(s(y), x), z)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( c(x1, x2) ) =
/0\
\0/
+
/01\
\10/
·x1+
/10\
\01/
·x2

M( s(x1) ) =
/0\
\0/
+
/00\
\10/
·x1

M( 0 ) =
/1\
\0/

Tuple symbols:
M( H(x1, x2) ) = 0+
[1,0]
·x1+
[0,1]
·x2


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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: