YES Termination w.r.t. Q proof of Beerendonk_07_5.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 7 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 QDPQMonotonicMRRProof (⇔, 28 ms)
        ↳20 QDP
        ↳21 QDPQMonotonicMRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 10 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES

(0) Obligation:

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

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))

The signature Sigma is {cond}

(2) Obligation:

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

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

(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:

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))
COND(true, x) → AND(even(x), gr(x, 0))
COND(true, x) → EVEN(x)
COND(true, x) → GR(x, 0)
COND(true, x) → P(x)
EVEN(s(s(x))) → EVEN(x)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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:

EVEN(s(s(x))) → EVEN(x)

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

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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].

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

(11) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

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:

  • EVEN(s(s(x))) → EVEN(x)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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:

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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].

cond(true, x0)

(18) Obligation:

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

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

(19) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented rules of the TRS R:

p(s(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(COND(x1, x2)) = 2·x2   
POL(and(x1, x2)) = 2   
POL(even(x1)) = 2   
POL(false) = 0   
POL(gr(x1, x2)) = 2·x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + 2·x1   
POL(true) = 0   

(20) Obligation:

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

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

(21) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented rules of the TRS R:

and(x, false) → false
and(false, x) → false

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(COND(x1, x2)) = 2·x1   
POL(and(x1, x2)) = 1   
POL(even(x1)) = x1   
POL(false) = 0   
POL(gr(x1, x2)) = 2·x1   
POL(p(x1)) = 2 + x1   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

(22) Obligation:

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

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
p(0) → 0

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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.


COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
COND(x1, x2)  =  COND(x1, x2)
true  =  true
and(x1, x2)  =  x2
gr(x1, x2)  =  x1
p(x1)  =  p
0  =  0
false  =  false
s(x1)  =  s

Knuth-Bendix order [KBO] with precedence:
p > 0 > false

and weight map:

s=3
0=1
true=2
p=1
COND_2=1
false=1

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

gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
p(0) → 0

(24) Obligation:

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
p(0) → 0

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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