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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)


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

MOD(x, y) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
MOD(x, y) → ISZERO(y)
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
MINUS(s(x), s(y)) → MINUS(x, y)
IF_MOD(false, true, x, y, z) → MOD(z, y)
MOD(x, y) → LE(y, x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

MOD(x, y) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
MOD(x, y) → ISZERO(y)
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
MINUS(s(x), s(y)) → MINUS(x, y)
IF_MOD(false, true, x, y, z) → MOD(z, y)
MOD(x, y) → LE(y, x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

MOD(x, y) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
MOD(x, y) → ISZERO(y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF_MOD(false, true, x, y, z) → MOD(z, y)
MOD(x, y) → LE(y, x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

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.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

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.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP

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

MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, x1, x2)

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