(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)
A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__DIV(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(true, X, Y) → MARK(X)
A__IF(false, X, Y) → MARK(Y)
MARK(minus(X1, X2)) → A__MINUS(X1, X2)
MARK(geq(X1, X2)) → A__GEQ(X1, X2)
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
The TRS R consists of the following rules:
a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)
The TRS R consists of the following rules:
a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.