YES Termination proof of ../tpdb/TRS/CSR_Maude/my-nat/MYNAT_nosorts.trs
Termination of the following Term Rewriting System could be proven:

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

and(tt, X) → X
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)

The replacement map contains the following entries:

and: {1}
tt: empty set
plus: {1, 2}
0: empty set
s: {1}
x: {1, 2}


↳ CSR
  ↳ Zantema-Transformation

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

and(tt, X) → X
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)

The replacement map contains the following entries:

and: {1}
tt: empty set
plus: {1, 2}
0: empty set
s: {1}
x: {1, 2}

We applied the Zantema [34] to transform the context-sensitive TRS to an usual TRS.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence

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

and(tt, X) → a(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
a(x) → x

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof

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

and(tt, X) → a(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
a(x) → x

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)


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

X(N, s(M)) → X(N, M)
X(N, s(M)) → PLUS(x(N, M), N)
PLUS(N, s(M)) → PLUS(N, M)
AND(tt, X) → A(X)

The TRS R consists of the following rules:

and(tt, X) → a(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
a(x) → x

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)

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

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof

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

X(N, s(M)) → X(N, M)
X(N, s(M)) → PLUS(x(N, M), N)
PLUS(N, s(M)) → PLUS(N, M)
AND(tt, X) → A(X)

The TRS R consists of the following rules:

and(tt, X) → a(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
a(x) → x

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP

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

PLUS(N, s(M)) → PLUS(N, M)

The TRS R consists of the following rules:

and(tt, X) → a(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
a(x) → x

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)

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.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                  ↳ QDP

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

PLUS(N, s(M)) → PLUS(N, M)

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

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)

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.

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)



↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ QDPSizeChangeProof
                  ↳ QDP

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

PLUS(N, s(M)) → PLUS(N, M)

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: 



↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof

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

X(N, s(M)) → X(N, M)

The TRS R consists of the following rules:

and(tt, X) → a(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
a(x) → x

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)

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.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof

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

X(N, s(M)) → X(N, M)

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

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)

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.

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
a(x0)



↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ QDPSizeChangeProof

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

X(N, s(M)) → X(N, M)

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: