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:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(U11(X1, X2)) → MARK(X1)
A__U11(tt, N) → MARK(N)
A__U21(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(s(X)) → MARK(X)
MARK(plus(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(U21(X1, X2, X3)) → A__U21(mark(X1), X2, X3)
MARK(plus(X1, X2)) → MARK(X2)
A__U21(tt, M, N) → MARK(M)
A__AND(tt, X) → MARK(X)
A__U21(tt, M, N) → MARK(N)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U21(X1, X2, X3)) → MARK(X1)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
A__PLUS(N, s(M)) → A__U21(a__and(a__isNat(M), isNat(N)), M, N)
MARK(and(X1, X2)) → MARK(X1)
A__PLUS(N, s(M)) → A__ISNAT(M)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(isNat(X)) → A__ISNAT(X)
A__PLUS(N, s(M)) → A__AND(a__isNat(M), isNat(N))
A__PLUS(N, 0) → A__U11(a__isNat(N), N)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(U11(X1, X2)) → MARK(X1)
A__U11(tt, N) → MARK(N)
A__U21(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(s(X)) → MARK(X)
MARK(plus(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(U21(X1, X2, X3)) → A__U21(mark(X1), X2, X3)
MARK(plus(X1, X2)) → MARK(X2)
A__U21(tt, M, N) → MARK(M)
A__AND(tt, X) → MARK(X)
A__U21(tt, M, N) → MARK(N)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U21(X1, X2, X3)) → MARK(X1)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
A__PLUS(N, s(M)) → A__U21(a__and(a__isNat(M), isNat(N)), M, N)
MARK(and(X1, X2)) → MARK(X1)
A__PLUS(N, s(M)) → A__ISNAT(M)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(isNat(X)) → A__ISNAT(X)
A__PLUS(N, s(M)) → A__AND(a__isNat(M), isNat(N))
A__PLUS(N, 0) → A__U11(a__isNat(N), N)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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.


MARK(s(X)) → MARK(X)
MARK(U21(X1, X2, X3)) → A__U21(mark(X1), X2, X3)
MARK(U21(X1, X2, X3)) → MARK(X1)
A__PLUS(N, s(M)) → A__U21(a__and(a__isNat(M), isNat(N)), M, N)
A__PLUS(N, s(M)) → A__ISNAT(M)
A__PLUS(N, s(M)) → A__AND(a__isNat(M), isNat(N))
The remaining pairs can at least be oriented weakly.

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(U11(X1, X2)) → MARK(X1)
A__U11(tt, N) → MARK(N)
A__U21(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(plus(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(plus(X1, X2)) → MARK(X2)
A__U21(tt, M, N) → MARK(M)
A__AND(tt, X) → MARK(X)
A__U21(tt, M, N) → MARK(N)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(isNat(X)) → A__ISNAT(X)
A__PLUS(N, 0) → A__U11(a__isNat(N), N)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(A__AND(x1, x2)) = x2   
POL(A__ISNAT(x1)) = 0   
POL(A__PLUS(x1, x2)) = x1 + x2   
POL(A__U11(x1, x2)) = x2   
POL(A__U21(x1, x2, x3)) = x2 + x3   
POL(MARK(x1)) = x1   
POL(U11(x1, x2)) = x1 + x2   
POL(U21(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(a__U11(x1, x2)) = x1 + x2   
POL(a__U21(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(a__and(x1, x2)) = x1 + x2   
POL(a__isNat(x1)) = 0   
POL(a__plus(x1, x2)) = x1 + x2   
POL(and(x1, x2)) = x1 + x2   
POL(isNat(x1)) = 0   
POL(mark(x1)) = x1   
POL(plus(x1, x2)) = x1 + x2   
POL(s(x1)) = 1 + x1   
POL(tt) = 0   

The following usable rules [17] were oriented:

mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
a__plus(N, 0) → a__U11(a__isNat(N), N)
mark(isNat(X)) → a__isNat(X)
a__isNat(s(V1)) → a__isNat(V1)
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__and(tt, X) → mark(X)
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__U11(tt, N) → mark(N)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__isNat(X) → isNat(X)
a__isNat(0) → tt
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(U11(X1, X2)) → MARK(X1)
A__U11(tt, N) → MARK(N)
A__U21(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(plus(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(plus(X1, X2)) → MARK(X2)
A__U21(tt, M, N) → MARK(M)
A__AND(tt, X) → MARK(X)
A__U21(tt, M, N) → MARK(N)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(isNat(X)) → A__ISNAT(X)
A__PLUS(N, 0) → A__U11(a__isNat(N), N)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ QDPOrderProof

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__U11(tt, N) → MARK(N)
MARK(U11(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(plus(X1, X2)) → MARK(X2)
A__AND(tt, X) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(isNat(X)) → A__ISNAT(X)
A__PLUS(N, 0) → A__U11(a__isNat(N), N)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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.


MARK(plus(X1, X2)) → MARK(X1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(plus(X1, X2)) → MARK(X2)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
A__PLUS(N, 0) → A__U11(a__isNat(N), N)
The remaining pairs can at least be oriented weakly.

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__U11(tt, N) → MARK(N)
MARK(U11(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__AND(tt, X) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(isNat(X)) → A__ISNAT(X)
Used ordering: Polynomial interpretation [25]:

POL(0) = 1   
POL(A__AND(x1, x2)) = x2   
POL(A__ISNAT(x1)) = 0   
POL(A__PLUS(x1, x2)) = x1 + x2   
POL(A__U11(x1, x2)) = x2   
POL(MARK(x1)) = x1   
POL(U11(x1, x2)) = x1 + x2   
POL(U21(x1, x2, x3)) = 1 + x2 + x3   
POL(a__U11(x1, x2)) = x1 + x2   
POL(a__U21(x1, x2, x3)) = 1 + x2 + x3   
POL(a__and(x1, x2)) = x1 + x2   
POL(a__isNat(x1)) = 0   
POL(a__plus(x1, x2)) = 1 + x1 + x2   
POL(and(x1, x2)) = x1 + x2   
POL(isNat(x1)) = 0   
POL(mark(x1)) = x1   
POL(plus(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = x1   
POL(tt) = 0   

The following usable rules [17] were oriented:

mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
a__plus(N, 0) → a__U11(a__isNat(N), N)
mark(isNat(X)) → a__isNat(X)
a__isNat(s(V1)) → a__isNat(V1)
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__and(tt, X) → mark(X)
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__U11(tt, N) → mark(N)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__isNat(X) → isNat(X)
a__isNat(0) → tt
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
QDP
                  ↳ QDPOrderProof

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

MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(U11(X1, X2)) → MARK(X1)
A__U11(tt, N) → MARK(N)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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.


MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U11(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__U11(tt, N) → MARK(N)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)
A__AND(tt, X) → MARK(X)
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 1   
POL(A__AND(x1, x2)) = x2   
POL(A__ISNAT(x1)) = 0   
POL(A__U11(x1, x2)) = x2   
POL(MARK(x1)) = x1   
POL(U11(x1, x2)) = 1 + x1 + x2   
POL(U21(x1, x2, x3)) = x2 + x3   
POL(a__U11(x1, x2)) = 1 + x1 + x2   
POL(a__U21(x1, x2, x3)) = x2 + x3   
POL(a__and(x1, x2)) = x1 + x2   
POL(a__isNat(x1)) = 0   
POL(a__plus(x1, x2)) = x1 + x2   
POL(and(x1, x2)) = x1 + x2   
POL(isNat(x1)) = 0   
POL(mark(x1)) = x1   
POL(plus(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   
POL(tt) = 0   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
QDP
                      ↳ DependencyGraphProof

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__U11(tt, N) → MARK(N)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
QDP
                          ↳ QDPOrderProof

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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.


MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)
A__AND(tt, X) → MARK(X)
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0   
POL(A__AND(x1, x2)) = x2   
POL(A__ISNAT(x1)) = 0   
POL(MARK(x1)) = x1   
POL(U11(x1, x2)) = 0   
POL(U21(x1, x2, x3)) = 0   
POL(a__U11(x1, x2)) = 0   
POL(a__U21(x1, x2, x3)) = 0   
POL(a__and(x1, x2)) = 0   
POL(a__isNat(x1)) = 0   
POL(a__plus(x1, x2)) = 0   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(isNat(x1)) = 0   
POL(mark(x1)) = 0   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = 0   
POL(tt) = 0   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ QDPOrderProof
QDP
                              ↳ QDPOrderProof

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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.


A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__AND(tt, X) → MARK(X)
The remaining pairs can at least be oriented weakly.

A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(A__AND(x1, x2)) = 1 + x2   
POL(A__ISNAT(x1)) = x1   
POL(MARK(x1)) = x1   
POL(U11(x1, x2)) = 0   
POL(U21(x1, x2, x3)) = 0   
POL(a__U11(x1, x2)) = 0   
POL(a__U21(x1, x2, x3)) = 0   
POL(a__and(x1, x2)) = 0   
POL(a__isNat(x1)) = 0   
POL(a__plus(x1, x2)) = 0   
POL(and(x1, x2)) = 0   
POL(isNat(x1)) = x1   
POL(mark(x1)) = 0   
POL(plus(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = x1   
POL(tt) = 0   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ QDPOrderProof
                            ↳ QDP
                              ↳ QDPOrderProof
QDP
                                  ↳ DependencyGraphProof

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

A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(s(V1)) → A__ISNAT(V1)
MARK(isNat(X)) → A__ISNAT(X)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ QDPOrderProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
QDP
                                      ↳ UsableRulesProof

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

A__ISNAT(s(V1)) → A__ISNAT(V1)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ QDPOrderProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ UsableRulesProof
QDP
                                          ↳ QDPSizeChangeProof

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

A__ISNAT(s(V1)) → A__ISNAT(V1)

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: