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 DependencyGraphProof (⇔)
    ↳24 QDP
    ↳25 Rewriting (⇔)
    ↳26 QDP
    ↳27 Rewriting (⇔)
    ↳28 QDP
    ↳29 DependencyGraphProof (⇔)
    ↳30 QDP
    ↳31 Rewriting (⇔)
    ↳32 QDP
    ↳33 Narrowing (⇔)
    ↳34 QDP
    ↳35 DependencyGraphProof (⇔)
    ↳36 QDP
    ↳37 UsableRulesProof (⇔)
    ↳38 QDP
    ↳39 Rewriting (⇔)
    ↳40 QDP
    ↳41 Rewriting (⇔)
    ↳42 QDP
    ↳43 Narrowing (⇔)
    ↳44 QDP
    ↳45 DependencyGraphProof (⇔)
    ↳46 QDP
    ↳47 UsableRulesProof (⇔)
    ↳48 QDP
    ↳49 QReductionProof (⇔)
    ↳50 QDP
    ↳51 Rewriting (⇔)
    ↳52 QDP
    ↳53 UsableRulesProof (⇔)
    ↳54 QDP
    ↳55 QReductionProof (⇔)
    ↳56 QDP
    ↳57 Rewriting (⇔)
    ↳58 QDP
    ↳59 UsableRulesProof (⇔)
    ↳60 QDP
    ↳61 QReductionProof (⇔)
    ↳62 QDP
    ↳63 QDPSizeChangeProof (⇔)
    ↳64 TRUE

(0) Obligation:

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

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

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

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

The TRS R 2 is

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

The signature Sigma is {cond}

(2) Obligation:

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

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

The set Q consists of the following terms:

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

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

cond(true, x0, x1)

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(19) Narrowing (EQUIVALENT transformation)

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

COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

(20) Obligation:

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

COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(21) Rewriting (EQUIVALENT transformation)

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

COND(true, 0, y1) → COND(false, p(0), p(y1))

(22) Obligation:

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, 0, y1) → COND(false, p(0), p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) Obligation:

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(25) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1)) at position [1] we obtained the following new rules [LPAR04]:

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

(26) Obligation:

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

COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(27) Rewriting (EQUIVALENT transformation)

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

COND(true, y0, 0) → COND(false, p(y0), p(0))

(28) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, 0) → COND(false, p(y0), p(0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(31) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)

(32) Obligation:

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(33) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1)) at position [0] we obtained the following new rules [LPAR04]:

COND(true, s(y0), 0) → COND(and(true, false), y0, p(0))
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

(34) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), 0) → COND(and(true, false), y0, p(0))
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(37) 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.

(38) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(39) Rewriting (EQUIVALENT transformation)

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

COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0)))

(40) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(41) Rewriting (EQUIVALENT transformation)

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

(42) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(43) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0) at position [0] we obtained the following new rules [LPAR04]:

COND(true, 0, s(y1)) → COND(and(false, true), p(0), y1)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

(44) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, 0, s(y1)) → COND(and(false, true), p(0), y1)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(47) 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.

(48) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

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

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

(49) 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))

(50) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

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

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

(51) Rewriting (EQUIVALENT transformation)

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

COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

(52) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

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

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

(53) 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.

(54) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

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

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

(55) 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].

and(true, true)
and(x0, false)
and(false, x0)

(56) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(57) Rewriting (EQUIVALENT transformation)

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

COND(true, s(x0), s(y1)) → COND(true, x0, y1)

(58) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(59) 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.

(60) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, 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.

(61) 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))

(62) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

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

(63) 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:

  • COND(true, s(y0), s(x0)) → COND(true, y0, x0)
    The graph contains the following edges 1 >= 1, 2 > 2, 3 > 3

(64) TRUE