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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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:

IF_GCD3(false, s1(x), s1(y)) -> GCD2(minus2(y, x), s1(x))
MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
IF_GCD3(true, s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(s1(x), y) -> LE2(s1(x), y)
LE2(s1(x), s1(y)) -> LE2(x, y)
GCD2(s1(x), s1(y)) -> LE2(y, x)
IF_GCD3(true, s1(x), s1(y)) -> GCD2(minus2(x, y), s1(y))
GCD2(s1(x), s1(y)) -> IF_GCD3(le2(y, x), s1(x), s1(y))
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)
IF_GCD3(false, s1(x), s1(y)) -> MINUS2(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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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:

IF_GCD3(false, s1(x), s1(y)) -> GCD2(minus2(y, x), s1(x))
MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
IF_GCD3(true, s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(s1(x), y) -> LE2(s1(x), y)
LE2(s1(x), s1(y)) -> LE2(x, y)
GCD2(s1(x), s1(y)) -> LE2(y, x)
IF_GCD3(true, s1(x), s1(y)) -> GCD2(minus2(x, y), s1(y))
GCD2(s1(x), s1(y)) -> IF_GCD3(le2(y, x), s1(x), s1(y))
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)
IF_GCD3(false, s1(x), s1(y)) -> MINUS2(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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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 4 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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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)) = 3·x1 + 3·x2   
POL(s1(x1)) = 2 + 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)
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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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 interpretation [21]:

POL(0) = 2   
POL(IF_MINUS3(x1, x2, x3)) = 2·x1 + x2 + 3·x3   
POL(MINUS2(x1, x2)) = 2 + 3·x1 + 3·x2   
POL(false) = 2   
POL(le2(x1, x2)) = 1 + x1   
POL(s1(x1)) = 1 + 3·x1   
POL(true) = 2   

The following usable rules [14] were oriented:

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



↳ 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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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:

IF_GCD3(false, s1(x), s1(y)) -> GCD2(minus2(y, x), s1(x))
IF_GCD3(true, s1(x), s1(y)) -> GCD2(minus2(x, y), s1(y))
GCD2(s1(x), s1(y)) -> IF_GCD3(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)
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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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_GCD3(false, s1(x), s1(y)) -> GCD2(minus2(y, x), s1(x))
IF_GCD3(true, s1(x), s1(y)) -> GCD2(minus2(x, y), s1(y))
GCD2(s1(x), s1(y)) -> IF_GCD3(le2(y, x), s1(x), s1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 3   
POL(GCD2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(IF_GCD3(x1, x2, x3)) = 1 + 2·x2 + 2·x3   
POL(false) = 3   
POL(if_minus3(x1, x2, x3)) = 1 + x2   
POL(le2(x1, x2)) = 3 + 3·x1   
POL(minus2(x1, x2)) = 1 + x1   
POL(s1(x1)) = 3 + x1   
POL(true) = 2   

The following usable rules [14] were oriented:

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



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

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))
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, s1(x), s1(y)) -> gcd2(minus2(x, y), s1(y))
if_gcd3(false, s1(x), s1(y)) -> gcd2(minus2(y, x), 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.