YES Termination w.r.t. Q proof of SK90_4.60.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 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPOrderProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES

(0) Obligation:

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

(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:

MSORT(.(x, y)) → MIN(x, y)
MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
MSORT(.(x, y)) → DEL(min(x, y), .(x, y))
MIN(x, .(y, z)) → MIN(x, z)
MIN(x, .(y, z)) → MIN(y, z)
DEL(x, .(y, z)) → DEL(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DEL(x, .(y, z)) → DEL(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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:

DEL(x, .(y, z)) → DEL(x, z)

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

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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].

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

(11) Obligation:

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

DEL(x, .(y, z)) → DEL(x, z)

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:

  • DEL(x, .(y, z)) → DEL(x, z)
    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:

MIN(x, .(y, z)) → MIN(y, z)
MIN(x, .(y, z)) → MIN(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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:

MIN(x, .(y, z)) → MIN(y, z)
MIN(x, .(y, z)) → MIN(x, z)

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

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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].

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

(18) Obligation:

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

MIN(x, .(y, z)) → MIN(y, z)
MIN(x, .(y, z)) → MIN(x, z)

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

(19) 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:

  • MIN(x, .(y, z)) → MIN(y, z)
    The graph contains the following edges 2 > 1, 2 > 2

  • MIN(x, .(y, z)) → MIN(x, z)
    The graph contains the following edges 1 >= 1, 2 > 2

(20) YES

(21) Obligation:

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

MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(22) 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.

(23) Obligation:

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

MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))
del(x, nil) → nil

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(24) 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].

msort(nil)
msort(.(x0, x1))

(25) Obligation:

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

MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))
del(x, nil) → nil

The set Q consists of the following terms:

min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(26) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MSORT(x1)  =  x1
.(x1, x2)  =  .
del(x1, x2)  =  del
if(x1, x2, x3)  =  if

Knuth-Bendix order [KBO] with precedence:
del > if

and weight map:

.=2
del=1
if=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

(27) Obligation:

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

min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))
del(x, nil) → nil

The set Q consists of the following terms:

min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES