(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
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:
MINUS(n__0, Y) → 01
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
MINUS(n__s(X), n__s(Y)) → ACTIVATE(Y)
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
GEQ(n__s(X), n__s(Y)) → ACTIVATE(X)
GEQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
DIV(s(X), n__s(Y)) → GEQ(X, activate(Y))
DIV(s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))
DIV(s(X), n__s(Y)) → MINUS(X, activate(Y))
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__0) → 01
ACTIVATE(n__s(X)) → S(X)
The TRS R consists of the following rules:
minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
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 13 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
The TRS R consists of the following rules:
minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
The TRS R consists of the following rules:
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
s(X) → n__s(X)
0 → n__0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
The TRS R consists of the following rules:
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
s(X) → n__s(X)
0 → n__0
The set Q consists of the following terms:
activate(x0)
s(x0)
0
We have to consider all minimal (P,Q,R)-chains.
(10) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
Strictly oriented rules of the TRS R:
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
s(X) → n__s(X)
0 → n__0
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 1
POL(GEQ(x1, x2)) = x1 + x2
POL(activate(x1)) = 2 + x1
POL(n__0) = 0
POL(n__s(x1)) = 3 + x1
POL(s(x1)) = 4 + x1
(11) Obligation:
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:
activate(x0)
s(x0)
0
We have to consider all minimal (P,Q,R)-chains.
(12) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
The TRS R consists of the following rules:
minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
The TRS R consists of the following rules:
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
s(X) → n__s(X)
0 → n__0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
The TRS R consists of the following rules:
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
s(X) → n__s(X)
0 → n__0
The set Q consists of the following terms:
activate(x0)
s(x0)
0
We have to consider all minimal (P,Q,R)-chains.
(19) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
Strictly oriented rules of the TRS R:
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
s(X) → n__s(X)
0 → n__0
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 1
POL(MINUS(x1, x2)) = x1 + x2
POL(activate(x1)) = 2 + x1
POL(n__0) = 0
POL(n__s(x1)) = 3 + x1
POL(s(x1)) = 4 + x1
(20) Obligation:
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:
activate(x0)
s(x0)
0
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) TRUE
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DIV(s(X), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))
The TRS R consists of the following rules:
minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(24) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
DIV(s(X), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(DIV(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(minus(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(activate(x1)) = | | + | | · | x1 |
The following usable rules [FROCOS05] were oriented:
0 → n__0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
minus(n__0, Y) → 0
(25) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(27) TRUE