YES Termination w.r.t. Q proof of Waldmann_06_jwmatchb1.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 (⇔, 126 ms)
↳10 QDP
↳11 PisEmptyProof (⇔, 0 ms)
↳12 YES

(0) Obligation:

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

h(f(x, y)) → f(y, f(h(h(x)), a))

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(y, f(h(h(x)), a))

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(x))
H(f(x, y)) → H(x)

The TRS R consists of the following rules:

h(f(x, y)) → f(y, f(h(h(x)), a))

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(x)) at position [0] we obtained the following new rules [LPAR04]:

H(f(f(x0, x1), y1)) → H(f(x1, f(h(h(x0)), a))) → H(f(f(x0, x1), y1)) → H(f(x1, f(h(h(x0)), a)))

(6) Obligation:

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

H(f(x, y)) → H(x)
H(f(f(x0, x1), y1)) → H(f(x1, f(h(h(x0)), a)))

The TRS R consists of the following rules:

h(f(x, y)) → f(y, f(h(h(x)), a))

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(x) we obtained the following new rules [LPAR04]:

H(f(f(y_0, y_1), x1)) → H(f(y_0, y_1)) → H(f(f(y_0, y_1), x1)) → H(f(y_0, y_1))
H(f(f(f(y_0, y_1), y_2), x1)) → H(f(f(y_0, y_1), y_2)) → H(f(f(f(y_0, y_1), y_2), x1)) → H(f(f(y_0, y_1), y_2))

(8) Obligation:

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

H(f(f(x0, x1), y1)) → H(f(x1, f(h(h(x0)), a)))
H(f(f(y_0, y_1), x1)) → H(f(y_0, y_1))
H(f(f(f(y_0, y_1), y_2), x1)) → H(f(f(y_0, y_1), y_2))

The TRS R consists of the following rules:

h(f(x, y)) → f(y, f(h(h(x)), a))

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(f(x0, x1), y1)) → H(f(x1, f(h(h(x0)), a)))
H(f(f(y_0, y_1), x1)) → H(f(y_0, y_1))
H(f(f(f(y_0, y_1), y_2), x1)) → H(f(f(y_0, y_1), y_2))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(H(x1)) = 1 +
[2,0]
·x1

POL(f(x1, x2)) =
/2\
\1/
 +
/03\
\01/
·x1 +
/01\
\01/
·x2

POL(h(x1)) =
/0\
\0/
 +
/03\
\10/
·x1

POL(a) =
/0\
\0/

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

h(f(x, y)) → f(y, f(h(h(x)), a))

(10) Obligation:

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

h(f(x, y)) → f(y, f(h(h(x)), a))

The set Q consists of the following terms:

h(f(x0, x1))

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) YES