(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, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, 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:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, s(y)))
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, s(x0))
minus(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
mod(s(x0), 0)
mod(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)
(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)
PLUS(x, s(y)) → PLUS(x, y)
MOD(x, s(y)) → HELP(x, s(y), 0)
HELP(x, s(y), c) → IF(le(c, x), x, s(y), c)
HELP(x, s(y), c) → LE(c, x)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
IF(true, x, s(y), c) → PLUS(c, s(y))
IF(false, x, s(y), c) → MINUS(x, minus(c, s(y)))
IF(false, x, s(y), c) → MINUS(c, s(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, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, s(y)))
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, s(x0))
minus(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
mod(s(x0), 0)
mod(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)
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 5 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(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, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, s(y)))
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, s(x0))
minus(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
mod(s(x0), 0)
mod(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(8) 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, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, s(y)))
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, s(x0))
minus(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
mod(s(x0), 0)
mod(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(9) 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, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, s(y)))
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, s(x0))
minus(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
mod(s(x0), 0)
mod(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HELP(x, s(y), c) → IF(le(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(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, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, s(y)))
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, s(x0))
minus(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
mod(s(x0), 0)
mod(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.