YES Termination w.r.t. Q proof of AProVE_04_IJCAR_26a.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 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 MNOCProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 MRRProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 DependencyGraphProof (⇔, 0 ms)
        ↳18 TRUE
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 QDPOrderProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPOrderProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 TransformationProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 DependencyGraphProof (⇔, 0 ms)
        ↳32 TRUE
    ↳33 QDP
        ↳34 QDPSizeChangeProof (⇔, 0 ms)
        ↳35 YES

(0) Obligation:

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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:

PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
PLUS(x, s(y)) → PLUS(x, p(s(y)))
PLUS(x, s(y)) → P(s(y))
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
DIV(x, y) → QUOT(x, y, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(div(x, y), z) → DIV(x, times(y, z))
DIV(div(x, y), z) → TIMES(y, z)
EQ(s(x), s(y)) → EQ(x, y)
DIVIDES(y, x) → EQ(x, times(div(x, y), y))
DIVIDES(y, x) → TIMES(div(x, y), y)
DIVIDES(y, x) → DIV(x, y)
PRIME(s(s(x))) → PR(s(s(x)), s(x))
PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
PR(x, s(s(y))) → DIVIDES(s(s(y)), x)
IF(false, x, y) → PR(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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 5 SCCs with 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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:

EQ(s(x), s(y)) → EQ(x, y)

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

(8) 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:

  • EQ(s(x), s(y)) → EQ(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(9) YES

(10) Obligation:

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

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → PLUS(x, y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

(11) 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.

(12) Obligation:

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

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → PLUS(x, y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

The TRS R consists of the following rules:

p(s(x)) → x

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

(13) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(14) Obligation:

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

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → PLUS(x, y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

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

(15) 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:

PLUS(s(x), y) → PLUS(x, y)

Strictly oriented rules of the TRS R:

p(s(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(PLUS(x1, x2)) = x1 + 2·x2   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + 2·x1   

(16) Obligation:

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

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

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

p(s(x0))

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(18) TRUE

(19) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

(20) 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.

(21) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

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

(22) 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:

  • TIMES(s(x), y) → TIMES(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(23) YES

(24) Obligation:

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

DIV(x, y) → QUOT(x, y, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(div(x, y), z) → DIV(x, times(y, z))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

(25) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


DIV(div(x, y), z) → DIV(x, times(y, z))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2)  =  x1
QUOT(x1, x2, x3)  =  x1
s(x1)  =  x1
div(x1, x2)  =  div(x1)

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

dummyConstant=1
div_1=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(26) Obligation:

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

DIV(x, y) → QUOT(x, y, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → DIV(x, s(z))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

(27) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


QUOT(s(x), s(y), z) → QUOT(x, y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2)  =  x1
QUOT(x1, x2, x3)  =  x1
s(x1)  =  s(x1)

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
dummyConstant=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(28) Obligation:

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

DIV(x, y) → QUOT(x, y, y)
QUOT(x, 0, s(z)) → DIV(x, s(z))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

(29) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule DIV(x, y) → QUOT(x, y, y) we obtained the following new rules [LPAR04]:

DIV(y_0, s(y_1)) → QUOT(y_0, s(y_1), s(y_1)) → DIV(y_0, s(y_1)) → QUOT(y_0, s(y_1), s(y_1))

(30) Obligation:

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

QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(y_0, s(y_1)) → QUOT(y_0, s(y_1), s(y_1))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

(31) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(32) TRUE

(33) Obligation:

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

PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
IF(false, x, y) → PR(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

(34) 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:

  • IF(false, x, y) → PR(x, y)
    The graph contains the following edges 2 >= 1, 3 >= 2

  • PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
    The graph contains the following edges 1 >= 2, 2 > 3

(35) YES