Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, x0, x1)
if_gcd(false, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) → LE(y, x)
IF_GCD(true, x, y) → GCD(minus(x, y), y)
IF_GCD(true, x, y) → MINUS(x, y)
IF_GCD(false, x, y) → GCD(minus(y, x), x)
IF_GCD(false, x, y) → MINUS(y, x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, x0, x1)
if_gcd(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, x0, x1)
if_gcd(false, x0, x1)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

s₁ > MINUS₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, x0, x1)
if_gcd(false, x0, x1)

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, x0, x1)
if_gcd(false, x0, x1)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2)  =  LE(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

s₁ > LE₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, x0, x1)
if_gcd(false, x0, x1)

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

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

GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
IF_GCD(true, x, y) → GCD(minus(x, y), y)
IF_GCD(false, x, y) → GCD(minus(y, x), x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, x0, x1)
if_gcd(false, x0, x1)

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