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:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

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:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(x1))


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

MINUS2(x, s1(y)) -> PRED1(minus2(x, y))
LE2(s1(x), s1(y)) -> LE2(x, y)
IF_MOD3(true, s1(x), s1(y)) -> MINUS2(x, y)
MOD2(s1(x), s1(y)) -> IF_MOD3(le2(y, x), s1(x), s1(y))
MINUS2(x, s1(y)) -> MINUS2(x, y)
IF_MOD3(true, s1(x), s1(y)) -> MOD2(minus2(x, y), s1(y))
MOD2(s1(x), s1(y)) -> LE2(y, x)

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(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)) -> PRED1(minus2(x, y))
LE2(s1(x), s1(y)) -> LE2(x, y)
IF_MOD3(true, s1(x), s1(y)) -> MINUS2(x, y)
MOD2(s1(x), s1(y)) -> IF_MOD3(le2(y, x), s1(x), s1(y))
MINUS2(x, s1(y)) -> MINUS2(x, y)
IF_MOD3(true, s1(x), s1(y)) -> MOD2(minus2(x, y), s1(y))
MOD2(s1(x), s1(y)) -> LE2(y, x)

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(x1))

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

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

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

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

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(x, s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(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
                ↳ QDPAfsSolverProof
              ↳ 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:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(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
QDP
                ↳ QDPAfsSolverProof

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

MOD2(s1(x), s1(y)) -> IF_MOD3(le2(y, x), s1(x), s1(y))
IF_MOD3(true, s1(x), s1(y)) -> MOD2(minus2(x, y), s1(y))

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF_MOD3(true, s1(x), s1(y)) -> MOD2(minus2(x, y), s1(y))
Used argument filtering: MOD2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
IF_MOD3(x1, x2, x3)  =  x2
minus2(x1, x2)  =  x1
pred1(x1)  =  x1
Used ordering: Quasi Precedence: trivial


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

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

MOD2(s1(x), s1(y)) -> IF_MOD3(le2(y, x), s1(x), s1(y))

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if_mod3(le2(y, x), s1(x), s1(y))
if_mod3(true, s1(x), s1(y)) -> mod2(minus2(x, y), s1(y))
if_mod3(false, s1(x), s1(y)) -> s1(x)

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
mod2(0, x0)
mod2(s1(x0), 0)
mod2(s1(x0), s1(x1))
if_mod3(true, s1(x0), s1(x1))
if_mod3(false, s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.