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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
MINUS2(s1(x), y) -> LE2(s1(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))
IF_MINUS3(false, s1(x), 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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
MINUS2(s1(x), y) -> LE2(s1(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))
IF_MINUS3(false, s1(x), 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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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 Order [17,21] with Interpretation:

POL( LE2(x1, x2) ) = max{0, 2x1 + 2x2 - 1}


POL( s1(x1) ) = 2x1 + 2



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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
IF_MINUS3(false, s1(x), 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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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_MINUS3(false, s1(x), y) -> MINUS2(x, y)
The remaining pairs can at least be oriented weakly.

MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = 2


POL( false ) = 0


POL( s1(x1) ) = x1 + 2


POL( 0 ) = 2


POL( le2(x1, x2) ) = 1


POL( IF_MINUS3(x1, ..., x3) ) = 2x2 + 2


POL( MINUS2(x1, x2) ) = 2x1 + 2



The following usable rules [14] were oriented: none



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

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

MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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.

↳ 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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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.


MOD2(s1(x), s1(y)) -> IF_MOD3(le2(y, x), s1(x), s1(y))
The remaining pairs can at least be oriented weakly.

IF_MOD3(true, s1(x), s1(y)) -> MOD2(minus2(x, y), s1(y))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( IF_MOD3(x1, ..., x3) ) = max{0, 2x2 - 2}


POL( true ) = 2


POL( minus2(x1, x2) ) = 2x1


POL( if_minus3(x1, ..., x3) ) = max{0, 2x2 - 1}


POL( false ) = 2


POL( 0 ) = max{0, -1}


POL( s1(x1) ) = 2x1 + 1


POL( le2(x1, x2) ) = 1


POL( MOD2(x1, x2) ) = max{0, 2x1 - 1}



The following usable rules [14] were oriented:

if_minus3(false, s1(x), y) -> s1(minus2(x, y))
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
minus2(0, y) -> 0
if_minus3(true, s1(x), y) -> 0



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

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

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)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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.