Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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
  ↳ Non-Overlap Check

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.

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:

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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(false, s1(x0), s1(x1))


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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(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:

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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(false, s1(x0), s1(x1))

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
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(false, s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


LE2(s1(X), s1(Y)) -> LE2(X, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE2(x1, x2)  =  LE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(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
                ↳ 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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(false, s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MINUS2(X, s1(Y)) -> MINUS2(X, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[MINUS1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(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

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))

The set Q consists of the following terms:

minus2(x0, s1(x1))
minus2(x0, 0)
pred1(s1(x0))
le2(s1(x0), s1(x1))
le2(s1(x0), 0)
le2(0, x0)
gcd2(0, x0)
gcd2(s1(x0), 0)
gcd2(s1(x0), s1(x1))
if3(true, s1(x0), s1(x1))
if3(false, s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.