YES Termination w.r.t. Q proof of Beerendonk_07_10.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 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 (⇔, 40 ms)
        ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QReductionProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPOrderProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 DependencyGraphProof (⇔, 0 ms)
        ↳35 TRUE

(0) Obligation:

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

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
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

neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))

The signature Sigma is {cond1, cond2}

(2) Obligation:

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

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
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:

COND1(true, x) → COND2(even(x), x)
COND1(true, x) → EVEN(x)
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(true, x) → NEQ(x, 0)
COND2(true, x) → DIV2(x)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(false, x) → NEQ(x, 0)
COND2(false, x) → P(x)
NEQ(s(x), s(y)) → NEQ(x, y)
EVEN(s(s(x))) → EVEN(x)
DIV2(s(s(x))) → DIV2(x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
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 3 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DIV2(s(s(x))) → DIV2(x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
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:

DIV2(s(s(x))) → DIV2(x)

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

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
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].

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

(11) Obligation:

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

DIV2(s(s(x))) → DIV2(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:

  • DIV2(s(s(x))) → DIV2(x)
    The graph contains the following edges 1 > 1

(13) YES

(14) 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:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
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:

EVEN(s(s(x))) → EVEN(x)

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

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
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].

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

(18) 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.

(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:

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

(20) YES

(21) Obligation:

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

COND2(true, x) → COND1(neq(x, 0), div2(x))
COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

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:

COND2(true, x) → COND1(neq(x, 0), div2(x))
COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

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].

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)

(25) Obligation:

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

COND2(true, x) → COND1(neq(x, 0), div2(x))
COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))

The set Q consists of the following terms:

neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(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.


COND2(true, x) → COND1(neq(x, 0), div2(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(COND1(x1, x2)) = [2] + x1 + [4]x2   
POL(COND2(x1, x2)) = [4] + [4]x2   
POL(div2(x1)) = [1/4]x1   
POL(even(x1)) = [2]   
POL(false) = [1/2]   
POL(neq(x1, x2)) = [1] + x1 + x2   
POL(p(x1)) = [1/4] + [1/4]x1   
POL(s(x1)) = [1] + [4]x1   
POL(true) = [2]   
The value of delta used in the strict ordering is 1.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

(27) Obligation:

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

COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))

The set Q consists of the following terms:

neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

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

(28) 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.

(29) Obligation:

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

COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)

The set Q consists of the following terms:

neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

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

(30) 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].

div2(0)
div2(s(0))
div2(s(s(x0)))

(31) Obligation:

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

COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)

The set Q consists of the following terms:

neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
p(0)
p(s(x0))

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

(32) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


COND2(false, x) → COND1(neq(x, 0), p(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = [1]   
POL(COND1(x1, x2)) = [1/4]x1 + [1/2]x2   
POL(COND2(x1, x2)) = x1 + [1/4]x2   
POL(even(x1)) = [1/4]x1   
POL(false) = [1/2]   
POL(neq(x1, x2)) = [1/4] + [1/4]x2   
POL(p(x1)) = [1/2] + [1/2]x1   
POL(s(x1)) = [4] + [4]x1   
POL(true) = 0   
The value of delta used in the strict ordering is 1/8.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x

(33) Obligation:

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

COND1(true, x) → COND2(even(x), x)

The TRS R consists of the following rules:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)

The set Q consists of the following terms:

neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
p(0)
p(s(x0))

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

(34) DependencyGraphProof (EQUIVALENT transformation)

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

(35) TRUE