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:

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(x0, s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)


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

MINUS2(x, s1(y)) -> MINUS2(x, p1(s1(y)))
MINUS2(x, s1(y)) -> P1(minus2(x, p1(s1(y))))
MINUS2(x, s1(y)) -> LE2(x, s1(y))
MINUS2(x, s1(y)) -> IF3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
LE2(s1(x), s1(y)) -> LE2(x, y)
MINUS2(x, s1(y)) -> P1(s1(y))

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(x0, s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

MINUS2(x, s1(y)) -> MINUS2(x, p1(s1(y)))
MINUS2(x, s1(y)) -> P1(minus2(x, p1(s1(y))))
MINUS2(x, s1(y)) -> LE2(x, s1(y))
MINUS2(x, s1(y)) -> IF3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
LE2(s1(x), s1(y)) -> LE2(x, y)
MINUS2(x, s1(y)) -> P1(s1(y))

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(x0, s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

LE2(s1(x), s1(y)) -> LE2(x, y)

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(x0, s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)

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


The following pairs can be strictly oriented and are deleted.


LE2(s1(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE2(x1, x2)  =  LE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(x0, s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

MINUS2(x, s1(y)) -> MINUS2(x, p1(s1(y)))

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(x)) -> x
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(x, s1(y)) -> if3(le2(x, s1(y)), 0, p1(minus2(x, p1(s1(y)))))
if3(true, x, y) -> x
if3(false, x, y) -> y

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(x0, s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)

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