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
  ↳ 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)

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)) -> 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)

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)) -> 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)

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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)

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, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS2(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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
          ↳ 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)

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 interpretation [21]:

POL(LE2(x1, x2)) = 2·x1 + x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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
          ↳ QDP
            ↳ QDPOrderProof

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)

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.


IF_MOD3(true, s1(x), s1(y)) -> MOD2(minus2(x, y), s1(y))
The remaining pairs can at least be oriented weakly.

MOD2(s1(x), s1(y)) -> IF_MOD3(le2(y, x), s1(x), s1(y))
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(IF_MOD3(x1, x2, x3)) = x2   
POL(MOD2(x1, x2)) = x1   
POL(false) = 0   
POL(le2(x1, x2)) = 1   
POL(minus2(x1, x2)) = x1   
POL(pred1(x1)) = x1   
POL(s1(x1)) = 1 + 2·x1   
POL(true) = 0   

The following usable rules [14] were oriented:

minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
pred1(s1(x)) -> x



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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.