(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))
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:
A(f, a(g, a(f, x))) → A(f, a(g, a(g, a(f, x))))
A(f, a(g, a(f, x))) → A(g, a(g, a(f, x)))
A(g, a(f, a(g, x))) → A(g, a(f, a(f, a(g, x))))
A(g, a(f, a(g, x))) → A(f, a(f, a(g, x)))
The TRS R consists of the following rules:
a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) UsableRulesReductionPairsProof (EQUIVALENT transformation)
First, we A-transformed [FROCOS05] the QDP-Problem.
Then we obtain the following A-transformed DP problem.
The pairs P are:
f1(g(f(x))) → f1(g(g(f(x))))
f1(g(f(x))) → g1(g(f(x)))
g1(f(g(x))) → g1(f(f(g(x))))
g1(f(g(x))) → f1(f(g(x)))
and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:
g(f(g(x))) → g(f(f(g(x))))
f(g(f(x))) → f(g(g(f(x))))
Q is empty.
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
The following rules are removed from R:
a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(f(x1)) = x1
POL(f1(x1)) = 2 + x1
POL(g(x1)) = x1
POL(g1(x1)) = 2 + x1
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f1(g(f(x))) → f1(g(g(f(x))))
f1(g(f(x))) → g1(g(f(x)))
g1(f(g(x))) → g1(f(f(g(x))))
g1(f(g(x))) → f1(f(g(x)))
The TRS R consists of the following rules:
g(f(g(x))) → g(f(f(g(x))))
f(g(f(x))) → f(g(g(f(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) RFCMatchBoundsDPProof (EQUIVALENT transformation)
Finiteness of the DP problem can be shown by a matchbound of 2.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:
f1(g(f(x))) → f1(g(g(f(x))))
f1(g(f(x))) → g1(g(f(x)))
g1(f(g(x))) → g1(f(f(g(x))))
g1(f(g(x))) → f1(f(g(x)))
To find matches we regarded all rules of R and P:
g(f(g(x))) → g(f(f(g(x))))
f(g(f(x))) → f(g(g(f(x))))
f1(g(f(x))) → f1(g(g(f(x))))
f1(g(f(x))) → g1(g(f(x)))
g1(f(g(x))) → g1(f(f(g(x))))
g1(f(g(x))) → f1(f(g(x)))
The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59
Node 40 is start node and node 41 is final node.
Those nodes are connected through the following edges:
- 40 to 42 labelled f1_1(0)
- 40 to 43 labelled g1_1(0)
- 40 to 45 labelled g1_1(0)
- 40 to 46 labelled f1_1(0)
- 41 to 41 labelled #_1(0)
- 42 to 43 labelled g_1(0)
- 43 to 44 labelled g_1(0)
- 43 to 51 labelled g_1(1)
- 44 to 41 labelled f_1(0)
- 44 to 48 labelled f_1(1)
- 45 to 46 labelled f_1(0)
- 46 to 47 labelled f_1(0)
- 46 to 48 labelled f_1(1)
- 47 to 41 labelled g_1(0)
- 47 to 51 labelled g_1(1)
- 48 to 49 labelled g_1(1)
- 49 to 50 labelled g_1(1)
- 49 to 51 labelled g_1(1)
- 49 to 54 labelled g_1(2)
- 50 to 41 labelled f_1(1)
- 50 to 48 labelled f_1(1)
- 51 to 52 labelled f_1(1)
- 52 to 53 labelled f_1(1)
- 52 to 48 labelled f_1(1)
- 52 to 57 labelled f_1(2)
- 53 to 41 labelled g_1(1)
- 53 to 51 labelled g_1(1)
- 54 to 55 labelled f_1(2)
- 55 to 56 labelled f_1(2)
- 56 to 49 labelled g_1(2)
- 57 to 58 labelled g_1(2)
- 58 to 59 labelled g_1(2)
- 59 to 52 labelled f_1(2)
(6) YES