YES Termination w.r.t. Q proof of Waldmann_06_jwmatchb2.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 (⇔, 0 ms)
↳4 QDP
↳5 TransformationProof (⇔, 0 ms)
↳6 QDP
↳7 TransformationProof (⇔, 0 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 168 ms)
↳10 QDP
↳11 UsableRulesProof (⇔, 0 ms)
↳12 QDP
↳13 QReductionProof (⇔, 0 ms)
↳14 QDP
↳15 QDPSizeChangeProof (⇔, 0 ms)
↳16 YES

(0) Obligation:

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

h(f(x, y)) → f(f(a, h(h(y))), x)

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:

h(f(x, y)) → f(f(a, h(h(y))), x)

The set Q consists of the following terms:

h(f(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:

H(f(x, y)) → H(h(y))
H(f(x, y)) → H(y)

The TRS R consists of the following rules:

h(f(x, y)) → f(f(a, h(h(y))), x)

The set Q consists of the following terms:

h(f(x0, x1))

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

(5) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule H(f(x, y)) → H(h(y)) at position [0] we obtained the following new rules [LPAR04]:

H(f(y0, f(x0, x1))) → H(f(f(a, h(h(x1))), x0)) → H(f(y0, f(x0, x1))) → H(f(f(a, h(h(x1))), x0))

(6) Obligation:

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

H(f(x, y)) → H(y)
H(f(y0, f(x0, x1))) → H(f(f(a, h(h(x1))), x0))

The TRS R consists of the following rules:

h(f(x, y)) → f(f(a, h(h(y))), x)

The set Q consists of the following terms:

h(f(x0, x1))

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

(7) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule H(f(x, y)) → H(y) we obtained the following new rules [LPAR04]:

H(f(x0, f(y_0, y_1))) → H(f(y_0, y_1)) → H(f(x0, f(y_0, y_1))) → H(f(y_0, y_1))
H(f(x0, f(y_0, f(y_1, y_2)))) → H(f(y_0, f(y_1, y_2))) → H(f(x0, f(y_0, f(y_1, y_2)))) → H(f(y_0, f(y_1, y_2)))

(8) Obligation:

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

H(f(y0, f(x0, x1))) → H(f(f(a, h(h(x1))), x0))
H(f(x0, f(y_0, y_1))) → H(f(y_0, y_1))
H(f(x0, f(y_0, f(y_1, y_2)))) → H(f(y_0, f(y_1, y_2)))

The TRS R consists of the following rules:

h(f(x, y)) → f(f(a, h(h(y))), x)

The set Q consists of the following terms:

h(f(x0, x1))

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

(9) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


H(f(y0, f(x0, x1))) → H(f(f(a, h(h(x1))), x0))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(H(x1)) = 0A +
[-I,1A,0A]
·x1

POL(f(x1, x2)) =
/3A\
|1A|
\3A/
 +
/3A2A0A\
|-I0A-I|
\-I0A-I/
·x1 +
/-I2A-I\
|-I0A-I|
\3A2A0A/
·x2

POL(a) =
/0A\
|1A|
\-I/

POL(h(x1)) =
/3A\
|0A|
\0A/
 +
/-I0A3A\
|-I0A-I|
\0A-I-I/
·x1

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

h(f(x, y)) → f(f(a, h(h(y))), x)

(10) Obligation:

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

H(f(x0, f(y_0, y_1))) → H(f(y_0, y_1))
H(f(x0, f(y_0, f(y_1, y_2)))) → H(f(y_0, f(y_1, y_2)))

The TRS R consists of the following rules:

h(f(x, y)) → f(f(a, h(h(y))), x)

The set Q consists of the following terms:

h(f(x0, x1))

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

(11) 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.

(12) Obligation:

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

H(f(x0, f(y_0, y_1))) → H(f(y_0, y_1))
H(f(x0, f(y_0, f(y_1, y_2)))) → H(f(y_0, f(y_1, y_2)))

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

h(f(x0, x1))

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

(13) 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].

h(f(x0, x1))

(14) Obligation:

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

H(f(x0, f(y_0, y_1))) → H(f(y_0, y_1))
H(f(x0, f(y_0, f(y_1, y_2)))) → H(f(y_0, f(y_1, y_2)))

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

(15) 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:

  • H(f(x0, f(y_0, y_1))) → H(f(y_0, y_1))
    The graph contains the following edges 1 > 1

  • H(f(x0, f(y_0, f(y_1, y_2)))) → H(f(y_0, f(y_1, y_2)))
    The graph contains the following edges 1 > 1

(16) YES