YES Termination w.r.t. Q proof of Rubio_04_gcd.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPOrderProof (⇔, 13 ms)
        ↳27 QDP
        ↳28 DependencyGraphProof (⇔, 0 ms)
        ↳29 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

(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:

MINUS(X, s(Y)) → PRED(minus(X, Y))
MINUS(X, s(Y)) → MINUS(X, Y)
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(true, s(X), s(Y)) → MINUS(X, Y)
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
IF(false, s(X), s(Y)) → MINUS(Y, X)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(X), s(Y)) → LE(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

LE(s(X), s(Y)) → LE(X, Y)

R is empty.
The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

(11) Obligation:

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

LE(s(X), s(Y)) → LE(X, Y)

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

(12) 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:

  • LE(s(X), s(Y)) → LE(X, Y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

MINUS(X, s(Y)) → MINUS(X, Y)

R is empty.
The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

(18) Obligation:

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

MINUS(X, s(Y)) → MINUS(X, Y)

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

(19) 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:

  • MINUS(X, s(Y)) → MINUS(X, Y)
    The graph contains the following edges 1 >= 1, 2 > 2

(20) YES

(21) Obligation:

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

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

(25) Obligation:

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

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( GCD(x1, x2) ) = max{0, 2x1 + x2 - 2}

POL( minus(x1, x2) ) = x1 + 1

POL( s(x1) ) = 2x1 + 2

POL( pred(x1) ) = max{0, x1 - 2}

POL( 0 ) = 2

POL( IF(x1, ..., x3) ) = max{0, x1 + 2x2 + x3 - 2}

POL( le(x1, x2) ) = 0

POL( false ) = 0

POL( true ) = 0


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

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
pred(s(X)) → X

(27) Obligation:

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

GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)

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

(28) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(29) TRUE