(0) Obligation:

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

f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, x)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, x)

The signature Sigma is {f}

(2) Obligation:

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

f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, x)

The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(s(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:

F(s(x), y) → F(x, s(s(x)))
F(x, s(s(y))) → F(y, x)

The TRS R consists of the following rules:

f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, x)

The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(s(x1)))

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

(5) 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.

(6) Obligation:

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

F(s(x), y) → F(x, s(s(x)))
F(x, s(s(y))) → F(y, x)

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

f(s(x0), x1)
f(x0, s(s(x1)))

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

(7) 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].

f(s(x0), x1)
f(x0, s(s(x1)))

(8) Obligation:

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

F(s(x), y) → F(x, s(s(x)))
F(x, s(s(y))) → F(y, x)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(F(x1, x2)) = (5/4)x1 + (1/4)x2   
POL(s(x1)) = 1/4 + (7/2)x1   
The value of delta used in the strict ordering is 1/32.
The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
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) TRUE