YES Termination proof of ../tpdb/TRS/CSR_Maude/peanoSimple/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))

The replacement map contains the following entries:

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


↳ 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))

The replacement map contains the following entries:

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

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

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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))
a(x) → x

Q is empty.

The following Q TRS is given: 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))
a(x) → x

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

and(tt, X) → a(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
a(x) → x
Used ordering:
Polynomial interpretation [25]:

POL(0) = 1   
POL(a(x1)) = 1 + x1   
POL(and(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(plus(x1, x2)) = 2 + x1 + 2·x2   
POL(s(x1)) = 1 + x1   
POL(tt) = 2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.