Termination w.r.t. Q proof of /tmp/tmp8LaKnv/18.trs

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

0 QTRS

↳1 AAECC Innermost (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 UsableRulesProof (⇔)

    ↳9 QDP

    ↳10 QReductionProof (⇔)

    ↳11 QDP

    ↳12 QDPSizeChangeProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 UsableRulesProof (⇔)

    ↳16 QDP

    ↳17 QReductionProof (⇔)

    ↳18 QDP

    ↳19 Narrowing (⇔)

    ↳20 QDP

    ↳21 Rewriting (⇔)

    ↳22 QDP

    ↳23 Rewriting (⇔)

    ↳24 QDP

    ↳25 Rewriting (⇔)

    ↳26 QDP

    ↳27 Rewriting (⇔)

    ↳28 QDP

    ↳29 Narrowing (⇔)

    ↳30 QDP

    ↳31 DependencyGraphProof (⇔)

    ↳32 QDP

    ↳33 Narrowing (⇔)

    ↳34 QDP

    ↳35 Rewriting (⇔)

    ↳36 QDP

    ↳37 Rewriting (⇔)

    ↳38 QDP

    ↳39 Rewriting (⇔)

    ↳40 QDP

    ↳41 Rewriting (⇔)

    ↳42 QDP

    ↳43 Narrowing (⇔)

    ↳44 QDP

    ↳45 DependencyGraphProof (⇔)

    ↳46 QDP

    ↳47 Narrowing (⇔)

    ↳48 QDP

    ↳49 DependencyGraphProof (⇔)

    ↳50 QDP

    ↳51 Rewriting (⇔)

    ↳52 QDP

    ↳53 Narrowing (⇔)

    ↳54 QDP

    ↳55 DependencyGraphProof (⇔)

    ↳56 QDP

    ↳57 Narrowing (⇔)

    ↳58 QDP

    ↳59 DependencyGraphProof (⇔)

    ↳60 QDP

    ↳61 Rewriting (⇔)

    ↳62 QDP

    ↳63 Narrowing (⇔)

    ↳64 QDP

    ↳65 DependencyGraphProof (⇔)

    ↳66 QDP

    ↳67 Rewriting (⇔)

    ↳68 QDP

    ↳69 UsableRulesProof (⇔)

    ↳70 QDP

    ↳71 Rewriting (⇔)

    ↳72 QDP

    ↳73 Narrowing (⇔)

    ↳74 QDP

    ↳75 DependencyGraphProof (⇔)

    ↳76 QDP

    ↳77 Narrowing (⇔)

    ↳78 QDP

    ↳79 DependencyGraphProof (⇔)

    ↳80 QDP

    ↳81 UsableRulesProof (⇔)

    ↳82 QDP

    ↳83 QReductionProof (⇔)

    ↳84 QDP

    ↳85 Rewriting (⇔)

    ↳86 QDP

    ↳87 UsableRulesProof (⇔)

    ↳88 QDP

    ↳89 QReductionProof (⇔)

    ↳90 QDP

    ↳91 Rewriting (⇔)

    ↳92 QDP

    ↳93 Instantiation (⇔)

    ↳94 QDP

    ↳95 DependencyGraphProof (⇔)

    ↳96 QDP

    ↳97 Instantiation (⇔)

    ↳98 QDP

    ↳99 DependencyGraphProof (⇔)

    ↳100 QDP

    ↳101 UsableRulesProof (⇔)

    ↳102 QDP

    ↳103 QReductionProof (⇔)

    ↳104 QDP

    ↳105 Instantiation (⇔)

    ↳106 QDP

    ↳107 DependencyGraphProof (⇔)

    ↳108 AND

      ↳109 QDP

        ↳110 QDPSizeChangeProof (⇔)

        ↳111 TRUE

      ↳112 QDP

        ↳113 QDPSizeChangeProof (⇔)

        ↳114 TRUE

(0) Obligation:

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

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

The TRS R 2 is

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

The signature Sigma is {cond1, cond2, cond3}

(2) Obligation:

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

cond1(true, x, y) → cond2(gr(x, 0), x, y)
cond2(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)
cond3(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3(false, x, y) → cond1(or(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)
or(false, false) → false
or(true, x) → true
or(x, true) → true
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)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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, 0), x, y)
COND1(true, x, y) → GR(x, 0)
COND2(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), p(x), y)
COND2(true, x, y) → OR(gr(x, 0), gr(y, 0))
COND2(true, x, y) → GR(x, 0)
COND2(true, x, y) → GR(y, 0)
COND2(true, x, y) → P(x)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND2(false, x, y) → GR(y, 0)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(true, x, y) → OR(gr(x, 0), gr(y, 0))
COND3(true, x, y) → GR(x, 0)
COND3(true, x, y) → GR(y, 0)
COND3(true, x, y) → P(y)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND3(false, x, y) → OR(gr(x, 0), gr(y, 0))
COND3(false, x, y) → GR(x, 0)
COND3(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, 0), x, y)
cond2(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)
cond3(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3(false, x, y) → cond1(or(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)
or(false, false) → false
or(true, x) → true
or(x, true) → true
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)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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, 0), x, y)
cond2(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)
cond3(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3(false, x, y) → cond1(or(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)
or(false, false) → false
or(true, x) → true
or(x, true) → true
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)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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) TRUE

(14) Obligation:

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

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

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, 0), x, y)
cond2(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)
cond3(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3(false, x, y) → cond1(or(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)
or(false, false) → false
or(true, x) → true
or(x, true) → true
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)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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) → COND1(or(gr(x, 0), gr(y, 0)), p(x), y)
COND1(true, x, y) → COND2(gr(x, 0), x, y)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(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
or(false, false) → false
or(true, x) → true
or(x, true) → true
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)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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)

(18) Obligation:

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

COND2(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), p(x), y)
COND1(true, x, y) → COND2(gr(x, 0), x, y)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(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
or(false, false) → false
or(true, x) → true
or(x, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

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

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

(19) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), p(x), y) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), p(0), y1)
COND2(true, s(x0), y1) → COND1(or(true, gr(y1, 0)), p(s(x0)), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), p(y0), s(x0))

(20) Obligation:

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

COND1(true, x, y) → COND2(gr(x, 0), x, y)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), p(0), y1)
COND2(true, s(x0), y1) → COND1(or(true, gr(y1, 0)), p(s(x0)), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), p(y0), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(21) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), p(0), y1) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)

(22) Obligation:

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

COND1(true, x, y) → COND2(gr(x, 0), x, y)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND2(true, s(x0), y1) → COND1(or(true, gr(y1, 0)), p(s(x0)), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), p(y0), s(x0))
COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(23) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), y1) → COND1(or(true, gr(y1, 0)), p(s(x0)), y1) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), y1) → COND1(true, p(s(x0)), y1)

(24) Obligation:

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

COND1(true, x, y) → COND2(gr(x, 0), x, y)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), p(y0), s(x0))
COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND2(true, s(x0), y1) → COND1(true, p(s(x0)), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(25) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), p(y0), s(x0)) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))

(26) Obligation:

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

COND1(true, x, y) → COND2(gr(x, 0), x, y)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND2(true, s(x0), y1) → COND1(true, p(s(x0)), y1)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(27) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), y1) → COND1(true, p(s(x0)), y1) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), y1) → COND1(true, x0, y1)

(28) Obligation:

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

COND1(true, x, y) → COND2(gr(x, 0), x, y)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(29) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND1(true, x, y) → COND2(gr(x, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)

(30) Obligation:

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

COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(31) DependencyGraphProof (EQUIVALENT transformation)

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

(32) Obligation:

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

COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(33) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(true, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, p(y)) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(or(true, gr(y1, 0)), s(x0), p(y1))
COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, p(0))
COND3(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), y0, p(s(x0)))

(34) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(or(true, gr(y1, 0)), s(x0), p(y1))
COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, p(0))
COND3(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(35) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, s(x0), y1) → COND1(or(true, gr(y1, 0)), s(x0), p(y1)) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))

(36) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, p(0))
COND3(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), y0, p(s(x0)))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(37) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, p(0)) at position [2] we obtained the following new rules [LPAR04]:

COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)

(38) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), y0, p(s(x0)))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(39) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, y0, s(x0)) → COND1(or(gr(y0, 0), true), y0, p(s(x0))) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, y0, s(x0)) → COND1(true, y0, p(s(x0)))

(40) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND3(true, y0, s(x0)) → COND1(true, y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(41) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, y0, s(x0)) → COND1(true, y0, p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

COND3(true, y0, s(x0)) → COND1(true, y0, x0)

(42) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND3(true, y0, s(x0)) → COND1(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(43) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(false, x, y) → COND3(gr(y, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

COND2(false, y0, 0) → COND3(false, y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))

(44) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(45) DependencyGraphProof (EQUIVALENT transformation)

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

(46) Obligation:

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

COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, y0, s(x0)) → COND1(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(47) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, x, y) → COND1(or(gr(x, 0), gr(y, 0)), x, y) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND3(false, s(x0), y1) → COND1(or(true, gr(y1, 0)), s(x0), y1)
COND3(false, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND3(false, y0, s(x0)) → COND1(or(gr(y0, 0), true), y0, s(x0))

(48) Obligation:

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

COND2(false, y0, 0) → COND3(false, y0, 0)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(false, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND3(false, s(x0), y1) → COND1(or(true, gr(y1, 0)), s(x0), y1)
COND3(false, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND3(false, y0, s(x0)) → COND1(or(gr(y0, 0), true), y0, s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(49) DependencyGraphProof (EQUIVALENT transformation)

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

(50) Obligation:

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

COND3(false, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(or(true, gr(y1, 0)), s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(false, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(51) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(false, s(x0), y1) → COND1(or(true, gr(y1, 0)), s(x0), y1) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, s(x0), y1) → COND1(true, s(x0), y1)

(52) Obligation:

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

COND3(false, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(false, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(53) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, 0, y1) → COND1(or(false, gr(y1, 0)), 0, y1) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, 0, 0) → COND1(or(false, false), 0, 0)
COND3(false, 0, s(x0)) → COND1(or(false, true), 0, s(x0))

(54) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(false, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND3(false, 0, 0) → COND1(or(false, false), 0, 0)
COND3(false, 0, s(x0)) → COND1(or(false, true), 0, s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(55) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(56) Obligation:

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

COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(57) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, y0, 0) → COND1(or(gr(y0, 0), false), y0, 0) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, 0, 0) → COND1(or(false, false), 0, 0)
COND3(false, s(x0), 0) → COND1(or(true, false), s(x0), 0)

(58) Obligation:

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

COND2(false, y0, 0) → COND3(false, y0, 0)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND3(false, 0, 0) → COND1(or(false, false), 0, 0)
COND3(false, s(x0), 0) → COND1(or(true, false), s(x0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(59) DependencyGraphProof (EQUIVALENT transformation)

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

(60) Obligation:

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

COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), 0) → COND1(or(true, false), s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(61) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(false, s(x0), 0) → COND1(or(true, false), s(x0), 0) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, s(x0), 0) → COND1(true, s(x0), 0)

(62) Obligation:

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

COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(63) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(true, y0, 0) → COND1(or(gr(y0, 0), false), p(y0), 0) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, 0, 0) → COND1(or(false, false), p(0), 0)
COND2(true, s(x0), 0) → COND1(or(true, false), p(s(x0)), 0)

(64) Obligation:

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

COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(true, 0, 0) → COND1(or(false, false), p(0), 0)
COND2(true, s(x0), 0) → COND1(or(true, false), p(s(x0)), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(65) DependencyGraphProof (EQUIVALENT transformation)

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

(66) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(or(true, false), p(s(x0)), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(67) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), 0) → COND1(or(true, false), p(s(x0)), 0) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), 0) → COND1(true, p(s(x0)), 0)

(68) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, p(s(x0)), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(69) 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.

(70) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, p(s(x0)), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(71) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), 0) → COND1(true, p(s(x0)), 0) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), 0) → COND1(true, x0, 0)

(72) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(73) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(true, y0, s(x0)) → COND1(true, p(y0), s(x0)) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND2(true, 0, s(y1)) → COND1(true, 0, s(y1))

(74) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND2(true, 0, s(y1)) → COND1(true, 0, s(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(75) DependencyGraphProof (EQUIVALENT transformation)

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

(76) Obligation:

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

COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(77) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(true, 0, y1) → COND1(or(false, gr(y1, 0)), 0, p(y1)) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, 0, 0) → COND1(or(false, false), 0, p(0))
COND3(true, 0, s(x0)) → COND1(or(false, true), 0, p(s(x0)))

(78) Obligation:

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

COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, 0) → COND1(or(false, false), 0, p(0))
COND3(true, 0, s(x0)) → COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(79) DependencyGraphProof (EQUIVALENT transformation)

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

(80) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(81) 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.

(82) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

or(x, true) → true
p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

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

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

(83) 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].

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

(84) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

or(x, true) → true
p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

or(false, false)
or(true, x0)
or(x0, true)
p(0)
p(s(x0))

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

(85) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, 0, s(x0)) → COND1(or(false, true), 0, p(s(x0))) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, 0, s(x0)) → COND1(true, 0, p(s(x0)))

(86) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(true, 0, p(s(x0)))

The TRS R consists of the following rules:

or(x, true) → true
p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

or(false, false)
or(true, x0)
or(x0, true)
p(0)
p(s(x0))

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

(87) 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.

(88) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(true, 0, p(s(x0)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

or(false, false)
or(true, x0)
or(x0, true)
p(0)
p(s(x0))

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

(89) 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].

or(false, false)
or(true, x0)
or(x0, true)

(90) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(true, 0, p(s(x0)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(91) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, 0, s(x0)) → COND1(true, 0, p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

COND3(true, 0, s(x0)) → COND1(true, 0, x0)

(92) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(true, 0, x0)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(93) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(false, y0, 0) → COND3(false, y0, 0) we obtained the following new rules [LPAR04]:

COND2(false, 0, 0) → COND3(false, 0, 0)

(94) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(false, s(x0), y1) → COND1(true, s(x0), y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(false, s(x0), 0) → COND1(true, s(x0), 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(true, 0, x0)
COND2(false, 0, 0) → COND3(false, 0, 0)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(95) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(96) Obligation:

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

COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(true, 0, x0)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(97) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(false, y0, s(x0)) → COND3(true, y0, s(x0)) we obtained the following new rules [LPAR04]:

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))

(98) Obligation:

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

COND3(true, s(x0), y1) → COND1(true, s(x0), p(y1))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND3(true, 0, s(x0)) → COND1(true, 0, x0)
COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(99) DependencyGraphProof (EQUIVALENT transformation)

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

(100) Obligation:

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

COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(true, 0, s(x0)) → COND1(true, 0, x0)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(101) 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.

(102) Obligation:

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

COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(true, 0, s(x0)) → COND1(true, 0, x0)

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

p(0)
p(s(x0))

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

(103) 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))

(104) Obligation:

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

COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND3(true, y0, s(x0)) → COND1(true, y0, x0)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(true, 0, s(x0)) → COND1(true, 0, x0)

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

(105) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND3(true, y0, s(x0)) → COND1(true, y0, x0) we obtained the following new rules [LPAR04]:

COND3(true, 0, s(z0)) → COND1(true, 0, z0)

(106) Obligation:

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

COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
COND3(true, 0, s(x0)) → COND1(true, 0, x0)

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

(107) DependencyGraphProof (EQUIVALENT transformation)

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

(108) Complex Obligation (AND)

(109) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND3(true, 0, s(x0)) → COND1(true, 0, x0)
COND1(true, 0, y1) → COND2(false, 0, y1)

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

(110) 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:

COND3(true, 0, s(x0)) → COND1(true, 0, x0)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
COND1(true, 0, y1) → COND2(false, 0, y1)
The graph contains the following edges 2 >= 2, 3 >= 3
COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
The graph contains the following edges 2 >= 2, 3 >= 3

(111) TRUE

(112) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(true, s(x0), 0) → COND1(true, x0, 0)
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))

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

(113) 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:

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3
COND2(true, s(x0), y1) → COND1(true, x0, y1)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
COND2(true, s(x0), 0) → COND1(true, x0, 0)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
COND2(true, s(x0), s(y1)) → COND1(true, x0, s(y1))
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3