YES Termination w.r.t. Q proof of Endrullis_06_pair3swap.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 MRRProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 28 ms)
↳6 QDP
↳7 PisEmptyProof (⇔, 0 ms)
↳8 YES

(0) Obligation:

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

p(a(a(x0)), p(x1, p(a(x2), x3))) → p(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

P(a(a(x0)), p(x1, p(a(x2), x3))) → P(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))
P(a(a(x0)), p(x1, p(a(x2), x3))) → P(a(a(b(x1))), p(a(a(x0)), x3))
P(a(a(x0)), p(x1, p(a(x2), x3))) → P(a(a(x0)), x3)

The TRS R consists of the following rules:

p(a(a(x0)), p(x1, p(a(x2), x3))) → p(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))

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

(3) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

P(a(a(x0)), p(x1, p(a(x2), x3))) → P(a(a(b(x1))), p(a(a(x0)), x3))
P(a(a(x0)), p(x1, p(a(x2), x3))) → P(a(a(x0)), x3)


Used ordering: Polynomial interpretation [POLO]:

POL(P(x1, x2)) = 2·x1 + x2   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(p(x1, x2)) = 2 + 2·x1 + x2   

(4) Obligation:

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

P(a(a(x0)), p(x1, p(a(x2), x3))) → P(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))

The TRS R consists of the following rules:

p(a(a(x0)), p(x1, p(a(x2), x3))) → p(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))

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

(5) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


P(a(a(x0)), p(x1, p(a(x2), x3))) → P(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(P(x1, x2)) = [4]x1 + x2   
POL(a(x1)) = [4] + [4]x1   
POL(b(x1)) = 0   
POL(p(x1, x2)) = [2]x1 + [1/2]x2   
The value of delta used in the strict ordering is 24.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

p(a(a(x0)), p(x1, p(a(x2), x3))) → p(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))

(6) Obligation:

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

p(a(a(x0)), p(x1, p(a(x2), x3))) → p(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) YES