(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(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:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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:
DIVISION(x, y) → DIV(x, y, 0)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
DIV(x, y, z) → LT(x, y)
DIV(x, y, z) → INC(z)
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
IF(false, x, s(y), z) → MINUS(x, s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
LT(s(x), s(y)) → LT(x, y)
INC(s(x)) → INC(x)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INC(s(x)) → INC(x)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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:
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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:
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
We have to consider all minimal (P,Q,R)-chains.