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 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE

(0) Obligation:

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

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

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

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

The TRS R 2 is

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

The signature Sigma is {cond}

(2) Obligation:

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

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

The set Q consists of the following terms:

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

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

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

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

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

(10) QReductionProof (EQUIVALENT transformation)

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

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

(11) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

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

(13) TRUE

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(17) QReductionProof (EQUIVALENT transformation)

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

cond(true, x0)

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(COND(x1, x2)) = (1/4)x1 + (3)x2   
POL(true) = 1/2   
POL(and(x1, x2)) = 1/4 + (1/2)x2   
POL(even(x1)) = 1/4 + (1/4)x1   
POL(gr(x1, x2)) = x1   
POL(0) = 0   
POL(p(x1)) = (1/4)x1   
POL(s(x1)) = 2 + (4)x1   
POL(false) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [FROCOS05] were oriented:

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

(20) Obligation:

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

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

The set Q consists of the following terms:

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

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

(21) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(22) TRUE