Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 Instantiation (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP
↳9 QDPOrderProof (⇔)
↳10 QDP
↳11 QDPOrderProof (⇔)
↳12 QDP
↳13 QDPOrderProof (⇔)
↳14 QDP
↳15 DependencyGraphProof (⇔)
↳16 QDP
↳17 Instantiation (⇔)
↳18 QDP
↳19 DependencyGraphProof (⇔)
↳20 QDP
↳21 UsableRulesProof (⇔)
↳22 QDP
↳23 QDPSizeChangeProof (⇔)
↳24 TRUE

(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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
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(n__div(n__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)
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__0) → 01
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__minus(X1, X2)) → MINUS(X1, X2)

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
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 1 SCC with 3 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__minus(X1, X2)) → MINUS(X1, X2)
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(Y)
IF(false, X, Y) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → GEQ(X, 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)) → 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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, X, Y) → ACTIVATE(Y) we obtained the following new rules [LPAR04]:

IF(false, n__s(n__div(n__minus(y_4, y_6), n__s(y_8))), n__0) → ACTIVATE(n__0)

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__minus(X1, X2)) → MINUS(X1, X2)
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → GEQ(X, 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)) → ACTIVATE(Y)
IF(false, n__s(n__div(n__minus(y_4, y_6), n__s(y_8))), n__0) → ACTIVATE(n__0)

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__minus(X1, X2)) → MINUS(X1, X2)
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)
DIV(s(X), n__s(Y)) → GEQ(X, 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)) → 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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) 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)) → ACTIVATE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(ACTIVATE(x1)) = 0A + 0A·x1

POL(n__s(x1)) = 0A + 0A·x1

POL(n__div(x1, x2)) = 5A + 0A·x1 + 1A·x2

POL(DIV(x1, x2)) = 5A + 0A·x1 + 1A·x2

POL(activate(x1)) = -I + 0A·x1

POL(s(x1)) = 0A + 0A·x1

POL(IF(x1, x2, x3)) = 0A + 0A·x1 + 0A·x2 + 0A·x3

POL(geq(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(n__minus(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(n__0) = 0A

POL(true) = 0A

POL(MINUS(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(GEQ(x1, x2)) = 5A + 0A·x1 + 0A·x2

POL(minus(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(if(x1, x2, x3)) = -I + -I·x1 + 0A·x2 + 1A·x3

POL(false) = 0A

POL(div(x1, x2)) = 5A + 0A·x1 + 1A·x2

POL(0) = 0A

The following usable rules [FROCOS05] were oriented:

activate(X) → X
activate(n__minus(X1, X2)) → minus(X1, X2)
if(false, X, Y) → activate(Y)
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
activate(n__div(X1, X2)) → div(activate(X1), X2)
if(true, X, Y) → activate(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
0 → n__0
s(X) → n__s(X)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
div(0, n__s(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))
minus(n__0, Y) → 0

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__minus(X1, X2)) → MINUS(X1, X2)
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)
DIV(s(X), n__s(Y)) → GEQ(X, 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)

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(n__s(X), n__s(Y)) → ACTIVATE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(ACTIVATE(x1)) = -I + 0A·x1

POL(n__s(x1)) = -I + 0A·x1

POL(n__div(x1, x2)) = 0A + 0A·x1 + 1A·x2

POL(DIV(x1, x2)) = 0A + 0A·x1 + 1A·x2

POL(activate(x1)) = -I + 0A·x1

POL(s(x1)) = -I + 0A·x1

POL(IF(x1, x2, x3)) = -I + -I·x1 + 0A·x2 + 0A·x3

POL(geq(x1, x2)) = 3A + -I·x1 + -I·x2

POL(n__minus(x1, x2)) = 0A + 0A·x1 + 1A·x2

POL(n__0) = 0A

POL(true) = 3A

POL(MINUS(x1, x2)) = 0A + 0A·x1 + 1A·x2

POL(GEQ(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(minus(x1, x2)) = 0A + 0A·x1 + 1A·x2

POL(if(x1, x2, x3)) = -I + -I·x1 + 0A·x2 + 0A·x3

POL(false) = 0A

POL(div(x1, x2)) = 0A + 0A·x1 + 1A·x2

POL(0) = 0A

The following usable rules [FROCOS05] were oriented:

activate(X) → X
activate(n__minus(X1, X2)) → minus(X1, X2)
if(false, X, Y) → activate(Y)
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
activate(n__div(X1, X2)) → div(activate(X1), X2)
if(true, X, Y) → activate(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
0 → n__0
s(X) → n__s(X)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
div(0, n__s(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))
minus(n__0, Y) → 0

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__minus(X1, X2)) → MINUS(X1, X2)
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
DIV(s(X), n__s(Y)) → GEQ(X, 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)

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__s(X)) → ACTIVATE(X)
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
DIV(s(X), n__s(Y)) → GEQ(X, 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(DIV(x1, x2)) = x1 + x2   
POL(GEQ(x1, x2)) = x1 + x2   
POL(IF(x1, x2, x3)) = x2   
POL(MINUS(x1, x2)) = x1   
POL(activate(x1)) = x1   
POL(div(x1, x2)) = x1 + x2   
POL(false) = 0   
POL(geq(x1, x2)) = 0   
POL(if(x1, x2, x3)) = x2 + x3   
POL(minus(x1, x2)) = x1   
POL(n__0) = 0   
POL(n__div(x1, x2)) = x1 + x2   
POL(n__minus(x1, x2)) = x1   
POL(n__s(x1)) = 1 + x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

activate(X) → X
activate(n__minus(X1, X2)) → minus(X1, X2)
if(false, X, Y) → activate(Y)
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
activate(n__div(X1, X2)) → div(activate(X1), X2)
if(true, X, Y) → activate(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
0 → n__0
s(X) → n__s(X)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
div(0, n__s(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))
minus(n__0, Y) → 0

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__minus(X1, X2)) → MINUS(X1, X2)

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, X, Y) → ACTIVATE(X) we obtained the following new rules [LPAR04]:

IF(true, n__s(n__div(n__minus(y_4, y_6), n__s(y_8))), n__0) → ACTIVATE(n__s(n__div(n__minus(y_4, y_6), n__s(y_8))))

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0)
ACTIVATE(n__div(X1, X2)) → DIV(activate(X1), X2)
ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
IF(true, n__s(n__div(n__minus(y_4, y_6), n__s(y_8))), n__0) → ACTIVATE(n__s(n__div(n__minus(y_4, y_6), n__s(y_8))))

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)

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(n__div(n__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)
div(X1, X2) → n__div(X1, X2)
minus(X1, X2) → n__minus(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__div(X1, X2)) → div(activate(X1), X2)
activate(n__minus(X1, X2)) → minus(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) 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.

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ACTIVATE(n__div(X1, X2)) → ACTIVATE(X1)
    The graph contains the following edges 1 > 1

(24) TRUE