YES Termination w.r.t. Q proof of AProVE_07_wiehe09.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 QDPOrderProof (⇔, 72 ms)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPOrderProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES
    ↳22 QDP
        ↳23 QDPOrderProof (⇔, 9 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(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:

MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(minus(x, y), z) → PLUS(y, z)
PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
PLUS(s(x), s(y)) → IF(gt(x, y), x, y)
PLUS(s(x), s(y)) → GT(x, y)
PLUS(s(x), s(y)) → IF(not(gt(x, y)), id(x), id(y))
PLUS(s(x), s(y)) → NOT(gt(x, y))
PLUS(s(x), s(y)) → ID(x)
PLUS(s(x), s(y)) → ID(y)
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(s(x), x) → IF(gt(x, x), id(x), id(x))
PLUS(s(x), x) → GT(x, x)
PLUS(s(x), x) → ID(x)
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(id(x), s(y)) → IF(gt(s(y), y), y, s(y))
PLUS(id(x), s(y)) → GT(s(y), y)
NOT(x) → IF(x, false, true)
GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(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 4 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(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:

GT(s(x), s(y)) → GT(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:

  • GT(s(x), s(y)) → GT(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), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(11) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

POL(s(x1)) = 0A + 2A·x1

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

POL(gt(x1, x2)) = -I + 0A·x1 + -I·x2

POL(id(x1)) = 0A + 1A·x1

POL(not(x1)) = 0A + 1A·x1

POL(zero) = 2A

POL(true) = 0A

POL(false) = 0A

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

gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)

(12) Obligation:

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

PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(13) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( PLUS(x1, x2) ) = x1 + 1

POL( if(x1, ..., x3) ) = 2x1 + x2 + x3

POL( gt(x1, x2) ) = max{0, -1}

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

POL( zero ) = 0

POL( true ) = 0

POL( false ) = 0

POL( id(x1) ) = x1 + 1


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

gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y

(14) Obligation:

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

PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(15) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( PLUS(x1, x2) ) = 2x1 + 2

POL( if(x1, ..., x3) ) = x1 + x2 + x3

POL( gt(x1, x2) ) = 0

POL( s(x1) ) = 2x1 + 1

POL( zero ) = 0

POL( true ) = 0

POL( false ) = 0

POL( id(x1) ) = x1


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

gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y

(16) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(17) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(18) YES

(19) Obligation:

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

MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(20) 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(minus(x, y), z) → MINUS(x, plus(y, z))
    The graph contains the following edges 1 > 1

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

(21) YES

(22) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

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

minus(x, 0) → x
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(s(x), s(y)) → minus(x, y)

(24) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) YES