(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
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:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
(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:
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
-1(s(x), s(y)) → -1(x, y)
GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
GCD(s(x), s(y)) → -1(s(max(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.