YES Termination w.r.t. Q proof of Strategy_removed_AG01_4.26.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 4 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPOrderProof (⇔, 31 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 TRUE

(0) Obligation:

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

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(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:

LE(s(x), s(y)) → LE(x, y)
MINUS(x, y) → IF(le(x, y), x, y)
MINUS(x, y) → LE(x, y)
IF(false, x, y) → MINUS(p(x), y)
IF(false, x, y) → P(x)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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:

LE(s(x), s(y)) → LE(x, y)

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

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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].

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(11) Obligation:

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

LE(s(x), s(y)) → LE(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:

  • LE(s(x), s(y)) → LE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

MINUS(x, y) → IF(le(x, y), x, y)
IF(false, x, y) → MINUS(p(x), y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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:

MINUS(x, y) → IF(le(x, y), x, y)
IF(false, x, y) → MINUS(p(x), y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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].

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(18) Obligation:

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

MINUS(x, y) → IF(le(x, y), x, y)
IF(false, x, y) → MINUS(p(x), y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(19) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IF(false, x, y) → MINUS(p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(IF(x1, x2, x3)) = [2]x1 + [1/4]x2   
POL(MINUS(x1, x2)) = [2] + x1   
POL(false) = [2]   
POL(le(x1, x2)) = [1] + [1/4]x1   
POL(p(x1)) = [1/4] + [1/4]x1   
POL(s(x1)) = [4] + [4]x1   
POL(true) = [1]   
The value of delta used in the strict ordering is 7/4.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
p(0) → 0
p(s(x)) → x

(20) Obligation:

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

MINUS(x, y) → IF(le(x, y), x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) TRUE