Termination w.r.t. Q of the following Term Rewriting System could be proven:
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.
↳ QTRS
↳ DependencyPairsProof
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.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → A(b(a(x)))
A(a(f(x, y))) → A(y)
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → A(b(a(b(a(x)))))
F(a(x), a(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(b(x), b(y)) → F(x, y)
A(a(f(x, y))) → A(x)
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → A(b(a(x)))
A(a(f(x, y))) → A(y)
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → A(b(a(b(a(x)))))
F(a(x), a(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(b(x), b(y)) → F(x, y)
A(a(f(x, y))) → A(x)
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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → A(y)
F(a(x), a(y)) → F(x, y)
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))))))
A(a(f(x, y))) → A(x)
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.
By using the rule removal processor [15] with the following polynomial ordering [25], 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(y)
A(a(f(x, y))) → A(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(F(x1, x2)) = 1 + x1 + x2
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(f(x1, x2)) = 1 + x1 + x2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(a(x), a(y)) → A(f(x, y))
F(a(x), a(y)) → 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)) → 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.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
The remaining pairs can at least be oriented weakly.
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)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( f(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( F(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
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))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(a(x), a(y)) → F(x, y)
F(a(x), a(y)) → A(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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
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.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
F(a(x), a(y)) → F(x, y)
F(b(x), b(y)) → F(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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(a(x), a(y)) → F(x, y)
The graph contains the following edges 1 > 1, 2 > 2
- F(b(x), b(y)) → F(x, y)
The graph contains the following edges 1 > 1, 2 > 2