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:

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
  ↳ 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

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:

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

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:

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

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 4 less nodes.

↳ 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

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.


LE2(s1(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LE2(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] 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:

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.
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
            ↳ QDPOrderProof

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

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.


MINUS2(x, s1(y)) -> MINUS2(x, p1(s1(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS2(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3


POL( p1(x1) ) = max{0, x1 - 3}



The following usable rules [14] were oriented:

p1(s1(x)) -> x



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

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

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.