Termination proof of /tmp/tmp0Eq1SV/MYNAT

Termination of the following Term Rewriting System could be proven:

Context-sensitive 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

  ↳ CSRInnermostProof

Context-sensitive 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}

The CSR is orthogonal. By [10] we can switch to innermost.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

Context-sensitive 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}

Innermost Strategy.

Using Improved CS-DPs we result in the following initial Q-CSDP problem.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {plus, s, x, PLUS, X} are replacing on all positions.
For all symbols f in {and, AND} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The ordinary context-sensitive dependency pairs DP_o are:

PLUS(N, s(M)) → PLUS(N, M)
X(N, s(M)) → PLUS(x(N, M), N)
X(N, s(M)) → X(N, M)

The collapsing dependency pairs are DP_c:

AND(tt, X) → X

The hidden terms of R are:
none

Every hiding context is built from:none

Hence, the new unhiding pairs DP_u are :

AND(tt, X) → U(X)

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 set Q consists of the following terms:

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

The approximation of the Context-Sensitive Dependency Graph contains 2 SCCs with 1 less node.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

                ↳ QCSDPSubtermProof

              ↳ QCSDP

Q-restricted context-sensitive dependency pair problem:
The symbols in {plus, s, x, PLUS} are replacing on all positions.
For all symbols f in {and} we have µ(f) = {1}.

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) → 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 set Q consists of the following terms:

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

We use the subterm processor [20].

The following pairs can be oriented strictly and are deleted.

PLUS(N, s(M)) → PLUS(N, M)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = x2

Subterm Order

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

                ↳ QCSDPSubtermProof

                  ↳ QCSDP

                    ↳ PIsEmptyProof

              ↳ QCSDP

Q-restricted context-sensitive dependency pair problem:
The symbols in {plus, s, x} are replacing on all positions.
For all symbols f in {and} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

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 set Q consists of the following terms:

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

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

              ↳ QCSDP

                ↳ QCSDPSubtermProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {plus, s, x, X} are replacing on all positions.
For all symbols f in {and} we have µ(f) = {1}.

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) → 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 set Q consists of the following terms:

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

We use the subterm processor [20].

The following pairs can be oriented strictly and are deleted.

X(N, s(M)) → X(N, M)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
X(x1, x2) = x2

Subterm Order

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

              ↳ QCSDP

                ↳ QCSDPSubtermProof

                  ↳ QCSDP

                    ↳ PIsEmptyProof

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 set Q consists of the following terms:

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

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