Termination w.r.t. Q of the given QTRS could be proven:

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
MINUS(s(x), y) → LE(s(x), y)
IF_MINUS(false, s(x), 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, s(x), s(y)) → GCD(minus(x, y), s(y))
IF_GCD(true, s(x), s(y)) → MINUS(x, y)
IF_GCD(false, s(x), s(y)) → GCD(minus(y, x), s(x))
IF_GCD(false, s(x), s(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(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(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, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) 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(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(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, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• LE(s(x), s(y)) → LE(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) Obligation:

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

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

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
false  =  false
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• IF_MINUS(false, s(x), y) → MINUS(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 > 1, 3 >= 2

• MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y) (allowed arguments on rhs = {2, 3})
  The graph contains the following edges 1 >= 2, 2 >= 3

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) Obligation:

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

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

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:

POL(0) = 0   
POL(false) = 1   
POL(if_minus(x₁, x₂, x₃)) = x₂   
POL(le(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   
POL(true) = 1   

From the DPs we obtained the following set of size-change graphs:

• GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y)) (allowed arguments on rhs = {2, 3})
  The graph contains the following edges 1 >= 2, 2 >= 3

• IF_GCD(true, s(x), s(y)) → GCD(minus(x, y), s(y)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 > 1, 3 >= 2

• IF_GCD(false, s(x), s(y)) → GCD(minus(y, x), s(x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 3 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05].

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

(13) TRUE