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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

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:

MINUS(x, y) → LT(y, x)
LT(s(x), s(y)) → LT(x, y)
MINUS(x, y) → HELP(lt(y, x), x, y)
HELP(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

MINUS(x, y) → LT(y, x)
LT(s(x), s(y)) → LT(x, y)
MINUS(x, y) → HELP(lt(y, x), x, y)
HELP(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

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

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

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

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.


LT(s(x), s(y)) → LT(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1 + (4)x_1   
POL(LT(x1, x2)) = (3)x_2   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none



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

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

MINUS(x, y) → HELP(lt(y, x), x, y)
HELP(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

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