YES Termination w.r.t. Q proof of Beerendonk_07_24.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 QDPQMonotonicMRRProof (⇔, 45 ms)
        ↳20 QDP
        ↳21 QDPQMonotonicMRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 12 ms)
        ↳24 QDP
        ↳25 QDPOrderProof (⇔, 25 ms)
        ↳26 QDP
        ↳27 UsableRulesProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 QReductionProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 NonInfProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 DependencyGraphProof (⇔, 0 ms)
        ↳34 TRUE

(0) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
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

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

The TRS R 2 is

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)

The signature Sigma is {cond1, cond2, cond3, cond4}

(2) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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, y) → COND2(gr(x, y), x, y)
COND1(true, x, y) → GR(x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND2(true, x, y) → GR(x, 0)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND2(false, x, y) → GR(y, 0)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(true, x, y) → GR(x, 0)
COND3(true, x, y) → P(x)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND3(false, x, y) → GR(x, 0)
COND3(false, x, y) → GR(y, 0)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(true, x, y) → GR(y, 0)
COND4(true, x, y) → P(y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND4(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND4(false, x, y) → GR(x, 0)
COND4(false, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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 13 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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:

GR(s(x), s(y)) → GR(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

(11) Obligation:

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

GR(s(x), s(y)) → GR(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:

  • GR(s(x), s(y)) → GR(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:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)

(18) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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) = 1   
POL(COND1(x1, x2, x3)) = x2 + 2·x3   
POL(COND2(x1, x2, x3)) = x2 + 2·x3   
POL(COND3(x1, x2, x3)) = 2·x1 + x2 + 2·x3   
POL(COND4(x1, x2, x3)) = x2 + 2·x3   
POL(and(x1, x2)) = 0   
POL(false) = 0   
POL(gr(x1, x2)) = 0   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

(20) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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(false, x) → false
and(x, false) → false

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(COND1(x1, x2, x3)) = 2·x1 + 2·x2   
POL(COND2(x1, x2, x3)) = 2 + 2·x2   
POL(COND3(x1, x2, x3)) = 2 + 2·x2   
POL(COND4(x1, x2, x3)) = 2 + 2·x2   
POL(and(x1, x2)) = 1   
POL(false) = 0   
POL(gr(x1, x2)) = 1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   
POL(true) = 1   

(22) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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.


COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(COND2(x1, x2, x3)) = 2A + -I·x1 + -I·x2 + 4A·x3

POL(true) = 2A

POL(COND3(x1, x2, x3)) = 0A + -I·x1 + -I·x2 + 4A·x3

POL(gr(x1, x2)) = -I + 0A·x1 + -I·x2

POL(0) = 0A

POL(p(x1)) = 0A + -I·x1

POL(false) = 0A

POL(COND1(x1, x2, x3)) = 0A + 0A·x1 + -I·x2 + 4A·x3

POL(and(x1, x2)) = -I + -I·x1 + 3A·x2

POL(COND4(x1, x2, x3)) = -I + 3A·x1 + -I·x2 + 4A·x3

POL(s(x1)) = 4A + -I·x1

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
p(0) → 0
and(true, true) → true

(24) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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.


COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(COND2(x1, x2, x3)) = -I + 0A·x1 + 2A·x2 + -I·x3

POL(true) = 3A

POL(COND3(x1, x2, x3)) = 2A + 1A·x1 + 2A·x2 + -I·x3

POL(gr(x1, x2)) = -I + 0A·x1 + -I·x2

POL(0) = 0A

POL(p(x1)) = 0A + -I·x1

POL(false) = 0A

POL(COND1(x1, x2, x3)) = -I + -I·x1 + 2A·x2 + -I·x3

POL(and(x1, x2)) = 0A + -I·x1 + -I·x2

POL(COND4(x1, x2, x3)) = -I + -I·x1 + 2A·x2 + -I·x3

POL(s(x1)) = 4A + 0A·x1

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
p(0) → 0
gr(s(x), s(y)) → gr(x, y)

(26) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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

(27) 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.

(28) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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

(29) 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].

p(0)
p(s(x0))

(30) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(31) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair COND2(true, x, y) → COND3(gr(x, 0), x, y) the following chains were created:
  • We consider the chain COND1(true, x4, x5) → COND2(gr(x4, x5), x4, x5), COND2(true, x6, x7) → COND3(gr(x6, 0), x6, x7) which results in the following constraint:
    (1)    (COND2(gr(x4, x5), x4, x5)=COND2(true, x6, x7) ⇒ COND2(true, x6, x7)≥COND3(gr(x6, 0), x6, x7))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gr(x4, x5)=true ⇒ COND2(true, x4, x5)≥COND3(gr(x4, 0), x4, x5))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x4, x5)=true which results in the following new constraints:
    (3)    (true=true ⇒ COND2(true, s(x63), 0)≥COND3(gr(s(x63), 0), s(x63), 0))

    (4)    (gr(x65, x64)=true∧(gr(x65, x64)=true ⇒ COND2(true, x65, x64)≥COND3(gr(x65, 0), x65, x64)) ⇒ COND2(true, s(x65), s(x64))≥COND3(gr(s(x65), 0), s(x65), s(x64)))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (COND2(true, s(x63), 0)≥COND3(gr(s(x63), 0), s(x63), 0))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gr(x65, x64)=true ⇒ COND2(true, x65, x64)≥COND3(gr(x65, 0), x65, x64)) with σ = [ ] which results in the following new constraint:
    (6)    (COND2(true, x65, x64)≥COND3(gr(x65, 0), x65, x64) ⇒ COND2(true, s(x65), s(x64))≥COND3(gr(s(x65), 0), s(x65), s(x64)))






For Pair COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) the following chains were created:
  • We consider the chain COND2(true, x12, x13) → COND3(gr(x12, 0), x12, x13), COND3(false, x14, x15) → COND1(and(gr(x14, 0), gr(x15, 0)), x14, x15) which results in the following constraint:
    (1)    (COND3(gr(x12, 0), x12, x13)=COND3(false, x14, x15) ⇒ COND3(false, x14, x15)≥COND1(and(gr(x14, 0), gr(x15, 0)), x14, x15))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (0=x66∧gr(x12, x66)=false ⇒ COND3(false, x12, x13)≥COND1(and(gr(x12, 0), gr(x13, 0)), x12, x13))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x12, x66)=false which results in the following new constraints:
    (3)    (false=false∧0=x67 ⇒ COND3(false, 0, x13)≥COND1(and(gr(0, 0), gr(x13, 0)), 0, x13))

    (4)    (gr(x70, x69)=false∧0=s(x69)∧(∀x71:gr(x70, x69)=false∧0=x69 ⇒ COND3(false, x70, x71)≥COND1(and(gr(x70, 0), gr(x71, 0)), x70, x71)) ⇒ COND3(false, s(x70), x13)≥COND1(and(gr(s(x70), 0), gr(x13, 0)), s(x70), x13))


    We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
    (5)    (COND3(false, 0, x13)≥COND1(and(gr(0, 0), gr(x13, 0)), 0, x13))


    We solved constraint (4) using rules (I), (II).




For Pair COND1(true, x, y) → COND2(gr(x, y), x, y) the following chains were created:
  • We consider the chain COND3(false, x26, x27) → COND1(and(gr(x26, 0), gr(x27, 0)), x26, x27), COND1(true, x28, x29) → COND2(gr(x28, x29), x28, x29) which results in the following constraint:
    (1)    (COND1(and(gr(x26, 0), gr(x27, 0)), x26, x27)=COND1(true, x28, x29) ⇒ COND1(true, x28, x29)≥COND2(gr(x28, x29), x28, x29))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (0=x74∧gr(x26, x74)=x72∧0=x75∧gr(x27, x75)=x73∧and(x72, x73)=true ⇒ COND1(true, x26, x27)≥COND2(gr(x26, x27), x26, x27))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on and(x72, x73)=true which results in the following new constraint:
    (3)    (true=true∧0=x74∧gr(x26, x74)=true∧0=x75∧gr(x27, x75)=true ⇒ COND1(true, x26, x27)≥COND2(gr(x26, x27), x26, x27))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (4)    (0=x74∧gr(x26, x74)=true∧0=x75∧gr(x27, x75)=true ⇒ COND1(true, x26, x27)≥COND2(gr(x26, x27), x26, x27))


    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on gr(x26, x74)=true which results in the following new constraints:
    (5)    (true=true∧0=0∧0=x75∧gr(x27, x75)=true ⇒ COND1(true, s(x77), x27)≥COND2(gr(s(x77), x27), s(x77), x27))

    (6)    (gr(x79, x78)=true∧0=s(x78)∧0=x75∧gr(x27, x75)=true∧(∀x80,x81:gr(x79, x78)=true∧0=x78∧0=x80∧gr(x81, x80)=true ⇒ COND1(true, x79, x81)≥COND2(gr(x79, x81), x79, x81)) ⇒ COND1(true, s(x79), x27)≥COND2(gr(s(x79), x27), s(x79), x27))


    We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
    (7)    (0=x75∧gr(x27, x75)=true ⇒ COND1(true, s(x77), x27)≥COND2(gr(s(x77), x27), s(x77), x27))


    We solved constraint (6) using rules (I), (II).We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on gr(x27, x75)=true which results in the following new constraints:
    (8)    (true=true∧0=0 ⇒ COND1(true, s(x77), s(x83))≥COND2(gr(s(x77), s(x83)), s(x77), s(x83)))

    (9)    (gr(x85, x84)=true∧0=s(x84)∧(∀x86:gr(x85, x84)=true∧0=x84 ⇒ COND1(true, s(x86), x85)≥COND2(gr(s(x86), x85), s(x86), x85)) ⇒ COND1(true, s(x77), s(x85))≥COND2(gr(s(x77), s(x85)), s(x77), s(x85)))


    We simplified constraint (8) using rules (I), (II) which results in the following new constraint:
    (10)    (COND1(true, s(x77), s(x83))≥COND2(gr(s(x77), s(x83)), s(x77), s(x83)))


    We solved constraint (9) using rules (I), (II).
  • We consider the chain COND4(false, x34, x35) → COND1(and(gr(x34, 0), gr(x35, 0)), x34, x35), COND1(true, x36, x37) → COND2(gr(x36, x37), x36, x37) which results in the following constraint:
    (1)    (COND1(and(gr(x34, 0), gr(x35, 0)), x34, x35)=COND1(true, x36, x37) ⇒ COND1(true, x36, x37)≥COND2(gr(x36, x37), x36, x37))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (0=x89∧gr(x34, x89)=x87∧0=x90∧gr(x35, x90)=x88∧and(x87, x88)=true ⇒ COND1(true, x34, x35)≥COND2(gr(x34, x35), x34, x35))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on and(x87, x88)=true which results in the following new constraint:
    (3)    (true=true∧0=x89∧gr(x34, x89)=true∧0=x90∧gr(x35, x90)=true ⇒ COND1(true, x34, x35)≥COND2(gr(x34, x35), x34, x35))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (4)    (0=x89∧gr(x34, x89)=true∧0=x90∧gr(x35, x90)=true ⇒ COND1(true, x34, x35)≥COND2(gr(x34, x35), x34, x35))


    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on gr(x34, x89)=true which results in the following new constraints:
    (5)    (true=true∧0=0∧0=x90∧gr(x35, x90)=true ⇒ COND1(true, s(x92), x35)≥COND2(gr(s(x92), x35), s(x92), x35))

    (6)    (gr(x94, x93)=true∧0=s(x93)∧0=x90∧gr(x35, x90)=true∧(∀x95,x96:gr(x94, x93)=true∧0=x93∧0=x95∧gr(x96, x95)=true ⇒ COND1(true, x94, x96)≥COND2(gr(x94, x96), x94, x96)) ⇒ COND1(true, s(x94), x35)≥COND2(gr(s(x94), x35), s(x94), x35))


    We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
    (7)    (0=x90∧gr(x35, x90)=true ⇒ COND1(true, s(x92), x35)≥COND2(gr(s(x92), x35), s(x92), x35))


    We solved constraint (6) using rules (I), (II).We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on gr(x35, x90)=true which results in the following new constraints:
    (8)    (true=true∧0=0 ⇒ COND1(true, s(x92), s(x98))≥COND2(gr(s(x92), s(x98)), s(x92), s(x98)))

    (9)    (gr(x100, x99)=true∧0=s(x99)∧(∀x101:gr(x100, x99)=true∧0=x99 ⇒ COND1(true, s(x101), x100)≥COND2(gr(s(x101), x100), s(x101), x100)) ⇒ COND1(true, s(x92), s(x100))≥COND2(gr(s(x92), s(x100)), s(x92), s(x100)))


    We simplified constraint (8) using rules (I), (II) which results in the following new constraint:
    (10)    (COND1(true, s(x92), s(x98))≥COND2(gr(s(x92), s(x98)), s(x92), s(x98)))


    We solved constraint (9) using rules (I), (II).




For Pair COND2(false, x, y) → COND4(gr(y, 0), x, y) the following chains were created:
  • We consider the chain COND1(true, x42, x43) → COND2(gr(x42, x43), x42, x43), COND2(false, x44, x45) → COND4(gr(x45, 0), x44, x45) which results in the following constraint:
    (1)    (COND2(gr(x42, x43), x42, x43)=COND2(false, x44, x45) ⇒ COND2(false, x44, x45)≥COND4(gr(x45, 0), x44, x45))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (gr(x42, x43)=false ⇒ COND2(false, x42, x43)≥COND4(gr(x43, 0), x42, x43))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x42, x43)=false which results in the following new constraints:
    (3)    (false=false ⇒ COND2(false, 0, x102)≥COND4(gr(x102, 0), 0, x102))

    (4)    (gr(x105, x104)=false∧(gr(x105, x104)=false ⇒ COND2(false, x105, x104)≥COND4(gr(x104, 0), x105, x104)) ⇒ COND2(false, s(x105), s(x104))≥COND4(gr(s(x104), 0), s(x105), s(x104)))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (COND2(false, 0, x102)≥COND4(gr(x102, 0), 0, x102))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gr(x105, x104)=false ⇒ COND2(false, x105, x104)≥COND4(gr(x104, 0), x105, x104)) with σ = [ ] which results in the following new constraint:
    (6)    (COND2(false, x105, x104)≥COND4(gr(x104, 0), x105, x104) ⇒ COND2(false, s(x105), s(x104))≥COND4(gr(s(x104), 0), s(x105), s(x104)))






For Pair COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) the following chains were created:
  • We consider the chain COND2(false, x56, x57) → COND4(gr(x57, 0), x56, x57), COND4(false, x58, x59) → COND1(and(gr(x58, 0), gr(x59, 0)), x58, x59) which results in the following constraint:
    (1)    (COND4(gr(x57, 0), x56, x57)=COND4(false, x58, x59) ⇒ COND4(false, x58, x59)≥COND1(and(gr(x58, 0), gr(x59, 0)), x58, x59))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (0=x106∧gr(x57, x106)=false ⇒ COND4(false, x56, x57)≥COND1(and(gr(x56, 0), gr(x57, 0)), x56, x57))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x57, x106)=false which results in the following new constraints:
    (3)    (false=false∧0=x107 ⇒ COND4(false, x56, 0)≥COND1(and(gr(x56, 0), gr(0, 0)), x56, 0))

    (4)    (gr(x110, x109)=false∧0=s(x109)∧(∀x111:gr(x110, x109)=false∧0=x109 ⇒ COND4(false, x111, x110)≥COND1(and(gr(x111, 0), gr(x110, 0)), x111, x110)) ⇒ COND4(false, x56, s(x110))≥COND1(and(gr(x56, 0), gr(s(x110), 0)), x56, s(x110)))


    We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
    (5)    (COND4(false, x56, 0)≥COND1(and(gr(x56, 0), gr(0, 0)), x56, 0))


    We solved constraint (4) using rules (I), (II).




To summarize, we get the following constraints P≥ for the following pairs.
  • COND2(true, x, y) → COND3(gr(x, 0), x, y)
    • (COND2(true, s(x63), 0)≥COND3(gr(s(x63), 0), s(x63), 0))
    • (COND2(true, x65, x64)≥COND3(gr(x65, 0), x65, x64) ⇒ COND2(true, s(x65), s(x64))≥COND3(gr(s(x65), 0), s(x65), s(x64)))

  • COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
    • (COND3(false, 0, x13)≥COND1(and(gr(0, 0), gr(x13, 0)), 0, x13))

  • COND1(true, x, y) → COND2(gr(x, y), x, y)
    • (COND1(true, s(x77), s(x83))≥COND2(gr(s(x77), s(x83)), s(x77), s(x83)))
    • (COND1(true, s(x92), s(x98))≥COND2(gr(s(x92), s(x98)), s(x92), s(x98)))

  • COND2(false, x, y) → COND4(gr(y, 0), x, y)
    • (COND2(false, 0, x102)≥COND4(gr(x102, 0), 0, x102))
    • (COND2(false, x105, x104)≥COND4(gr(x104, 0), x105, x104) ⇒ COND2(false, s(x105), s(x104))≥COND4(gr(s(x104), 0), s(x105), s(x104)))

  • COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
    • (COND4(false, x56, 0)≥COND1(and(gr(x56, 0), gr(0, 0)), x56, 0))




The constraints for P> respective Pbound are constructed from P≥ where we just replace every occurence of "t ≥ s" in P≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(COND1(x1, x2, x3)) = -1 + x1 + x2 + x3   
POL(COND2(x1, x2, x3)) = 1 + x1   
POL(COND3(x1, x2, x3)) = x1 - x2 + x3   
POL(COND4(x1, x2, x3)) = 1 + x1 + x2 - x3   
POL(and(x1, x2)) = x1 + x2   
POL(c) = -1   
POL(false) = 1   
POL(gr(x1, x2)) = 1   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
The following pairs are in Pbound:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
The following rules are usable:

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

(32) Obligation:

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

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

(33) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(34) TRUE