(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → 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:
p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(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(x, s(y)) → IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
MINUS(x, s(y)) → LE(x, s(y))
MINUS(x, s(y)) → P(minus(x, p(s(y))))
MINUS(x, s(y)) → MINUS(x, p(s(y)))
MINUS(x, s(y)) → P(s(y))
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(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 2 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:
p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(x, s(y)) → MINUS(x, p(s(y)))
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.