(0) Obligation:

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

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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:

MINUS(X, s(Y)) → PRED(minus(X, Y))
MINUS(X, s(Y)) → MINUS(X, Y)
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(true, s(X), s(Y)) → MINUS(X, Y)
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
IF(false, s(X), s(Y)) → MINUS(Y, X)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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 4 less nodes.

(6) Complex Obligation (AND)

(7) 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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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.


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)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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:

MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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.


MINUS(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)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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:

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

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

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