(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
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(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) → A(b(a(b(a(x)))))
A(a(f(x, y))) → A(b(a(x)))
A(a(f(x, y))) → A(x)
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → A(y)
F(a(x), a(y)) → A(f(x, y))
F(a(x), a(y)) → F(x, y)
F(b(x), b(y)) → F(x, y)
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(x), a(y)) → F(x, y)
F(b(x), b(y)) → F(x, y)
A(a(f(x, y))) → A(x)
A(a(f(x, y))) → A(y)
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) 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:
A(a(f(x, y))) → A(x)
A(a(f(x, y))) → A(y)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 1 + x1
POL(F(x1, x2)) = 2 + x1 + x2
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(f(x1, x2)) = 1 + x1 + x2
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(x), a(y)) → F(x, y)
F(b(x), b(y)) → F(x, y)
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(a(x), a(y)) → A(f(x, y))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :
| POL(F(x1, x2)) = | 1 | + | | · | x1 | + | | · | x2 |
| POL(f(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(a(x), a(y)) → a(f(x, y))
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(b(x), b(y)) → b(f(x, y))
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(x), a(y)) → F(x, y)
F(b(x), b(y)) → F(x, y)
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → F(x, y)
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → F(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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:
- F(b(x), b(y)) → F(x, y)
The graph contains the following edges 1 > 1, 2 > 2
- F(a(x), a(y)) → F(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(14) YES