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:
f(f(x, a), y) → f(y, f(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x, a), y) → f(y, 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:
F(f(x, a), y) → F(x, y)
F(f(x, a), y) → F(y, f(x, y))
The TRS R consists of the following rules:
f(f(x, a), y) → f(y, f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(x, a), y) → F(x, y)
F(f(x, a), y) → F(y, f(x, y))
The TRS R consists of the following rules:
f(f(x, a), y) → f(y, f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(f(x, a), y) → F(y, f(x, y)) we obtained the following new rules:
F(f(x0, a), f(y_0, a)) → F(f(y_0, a), f(x0, f(y_0, a)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(x, a), y) → F(x, y)
F(f(x0, a), f(y_0, a)) → F(f(y_0, a), f(x0, f(y_0, a)))
The TRS R consists of the following rules:
f(f(x, a), y) → f(y, f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(f(x, a), y) → F(x, y) we obtained the following new rules:
F(f(f(y_0, a), a), f(y_1, a)) → F(f(y_0, a), f(y_1, a))
F(f(f(y_0, a), a), x1) → F(f(y_0, a), x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(f(f(y_0, a), a), f(y_1, a)) → F(f(y_0, a), f(y_1, a))
F(f(x0, a), f(y_0, a)) → F(f(y_0, a), f(x0, f(y_0, a)))
F(f(f(y_0, a), a), x1) → F(f(y_0, a), x1)
The TRS R consists of the following rules:
f(f(x, a), y) → f(y, 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.
F(f(x0, a), f(y_0, a)) → F(f(y_0, a), f(x0, f(y_0, a)))
The remaining pairs can at least be oriented weakly.
F(f(f(y_0, a), a), f(y_1, a)) → F(f(y_0, a), f(y_1, a))
F(f(f(y_0, a), a), x1) → F(f(y_0, a), x1)
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:
f(f(x, a), y) → f(y, f(x, y))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
F(f(f(y_0, a), a), f(y_1, a)) → F(f(y_0, a), f(y_1, a))
F(f(f(y_0, a), a), x1) → F(f(y_0, a), x1)
The TRS R consists of the following rules:
f(f(x, a), y) → f(y, f(x, y))
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(f(f(y_0, a), a), f(y_1, a)) → F(f(y_0, a), f(y_1, a))
The graph contains the following edges 1 > 1, 2 >= 2
- F(f(f(y_0, a), a), x1) → F(f(y_0, a), x1)
The graph contains the following edges 1 > 1, 2 >= 2