Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DIFF(X, Y) → LEQ(X, Y)
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(diff(p(X), Y)))
ACTIVATE(n__0) → 01
ACTIVATE(n__s(X)) → S(X)
LEQ(s(X), s(Y)) → LEQ(X, Y)
IF(false, X, Y) → ACTIVATE(Y)
DIFF(X, Y) → P(X)
IF(true, X, Y) → ACTIVATE(X)
DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

DIFF(X, Y) → LEQ(X, Y)
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(diff(p(X), Y)))
ACTIVATE(n__0) → 01
ACTIVATE(n__s(X)) → S(X)
LEQ(s(X), s(Y)) → LEQ(X, Y)
IF(false, X, Y) → ACTIVATE(Y)
DIFF(X, Y) → P(X)
IF(true, X, Y) → ACTIVATE(X)
DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

DIFF(X, Y) → LEQ(X, Y)
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(diff(p(X), Y)))
ACTIVATE(n__0) → 01
LEQ(s(X), s(Y)) → LEQ(X, Y)
ACTIVATE(n__s(X)) → S(X)
IF(false, X, Y) → ACTIVATE(Y)
DIFF(X, Y) → P(X)
IF(true, X, Y) → ACTIVATE(X)
DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

LEQ(s(X), s(Y)) → LEQ(X, Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

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