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:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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:

IF3(false, s1(X), s1(Y)) -> MINUS2(Y, X)
IF3(true, s1(X), s1(Y)) -> GCD2(minus2(X, Y), s1(Y))
IF3(false, s1(X), s1(Y)) -> GCD2(minus2(Y, X), s1(X))
MINUS2(X, s1(Y)) -> PRED1(minus2(X, Y))
LE2(s1(X), s1(Y)) -> LE2(X, Y)
GCD2(s1(X), s1(Y)) -> IF3(le2(Y, X), s1(X), s1(Y))
GCD2(s1(X), s1(Y)) -> LE2(Y, X)
IF3(true, s1(X), s1(Y)) -> MINUS2(X, Y)
MINUS2(X, s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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:

IF3(false, s1(X), s1(Y)) -> MINUS2(Y, X)
IF3(true, s1(X), s1(Y)) -> GCD2(minus2(X, Y), s1(Y))
IF3(false, s1(X), s1(Y)) -> GCD2(minus2(Y, X), s1(X))
MINUS2(X, s1(Y)) -> PRED1(minus2(X, Y))
LE2(s1(X), s1(Y)) -> LE2(X, Y)
GCD2(s1(X), s1(Y)) -> IF3(le2(Y, X), s1(X), s1(Y))
GCD2(s1(X), s1(Y)) -> LE2(Y, X)
IF3(true, s1(X), s1(Y)) -> MINUS2(X, Y)
MINUS2(X, s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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)) = 2·x1 + 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:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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(X, s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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.


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
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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
          ↳ QDP
            ↳ QDPOrderProof

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

IF3(true, s1(X), s1(Y)) -> GCD2(minus2(X, Y), s1(Y))
IF3(false, s1(X), s1(Y)) -> GCD2(minus2(Y, X), s1(X))
GCD2(s1(X), s1(Y)) -> IF3(le2(Y, X), s1(X), s1(Y))

The TRS R consists of the following rules:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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.


IF3(true, s1(X), s1(Y)) -> GCD2(minus2(X, Y), s1(Y))
IF3(false, s1(X), s1(Y)) -> GCD2(minus2(Y, X), s1(X))
GCD2(s1(X), s1(Y)) -> IF3(le2(Y, X), s1(X), s1(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(GCD2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(IF3(x1, x2, x3)) = 2·x2 + 2·x3   
POL(false) = 0   
POL(le2(x1, x2)) = 0   
POL(minus2(x1, x2)) = x1   
POL(pred1(x1)) = x1   
POL(s1(x1)) = 2 + 2·x1   
POL(true) = 0   

The following usable rules [14] were oriented:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X



↳ 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:

minus2(X, s1(Y)) -> pred1(minus2(X, Y))
minus2(X, 0) -> X
pred1(s1(X)) -> X
le2(s1(X), s1(Y)) -> le2(X, Y)
le2(s1(X), 0) -> false
le2(0, Y) -> true
gcd2(0, Y) -> 0
gcd2(s1(X), 0) -> s1(X)
gcd2(s1(X), s1(Y)) -> if3(le2(Y, X), s1(X), s1(Y))
if3(true, s1(X), s1(Y)) -> gcd2(minus2(X, Y), s1(Y))
if3(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.