YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex49_GM04_Z.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 MNOCProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 MNOCProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 QDPOrderProof (⇔, 22 ms)
        ↳25 QDP
        ↳26 PisEmptyProof (⇔, 0 ms)
        ↳27 YES

(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: Knuth-Bendix order [KBO] with precedence:
GEQ2 > s1 > ns1 > 0 > activate1 > n0

and weight map:

n__0=1
0=1
activate_1=1
n__s_1=1
s_1=1
GEQ_2=0

The variable weight is 1

(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) YES

(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: Knuth-Bendix order [KBO] with precedence:
MINUS2 > s1 > ns1 > 0 > activate1 > n0

and weight map:

n__0=1
0=1
activate_1=1
n__s_1=1
s_1=1
MINUS_2=0

The variable weight is 1

(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) YES

(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,JAR06].


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: Combined order from the following AFS and order.
DIV(x1, x2)  =  x1
s(x1)  =  s
minus(x1, x2)  =  minus
0  =  0
n__0  =  n__0

Knuth-Bendix order [KBO] with precedence:
s > minus > 0 > n0

and weight map:

s=1
0=1
minus=1
n__0=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
0 → n__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) YES