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:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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:
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(a(a(0, x), y)) → C(c(0))
C(a(a(0, x), y)) → C(0)
C(c(c(y))) → C(a(y, 0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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:
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(a(a(0, x), y)) → C(c(0))
C(a(a(0, x), y)) → C(0)
C(c(c(y))) → C(a(y, 0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(a(a(0, x), y)) → C(c(0))
C(c(c(y))) → C(a(y, 0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(a(0, x), y)) → C(c(0)) at position [0] we obtained the following new rules:
C(a(a(0, y0), y1)) → C(0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(a(a(0, y0), y1)) → C(0)
C(c(c(y))) → C(a(y, 0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(c(c(y))) → C(a(y, 0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(c(y))) → C(c(a(y, 0))) at position [0] we obtained the following new rules:
C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))
C(c(c(y0))) → C(a(y0, 0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))
C(c(c(y))) → C(a(y, 0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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.
C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))
The remaining pairs can at least be oriented weakly.
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(a(y, 0))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( a(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
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:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(a(y, 0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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.
C(c(c(y))) → C(a(y, 0))
The remaining pairs can at least be oriented weakly.
C(a(a(0, x), y)) → C(c(c(0)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( a(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
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:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, x), y)) → C(c(c(0)))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(a(0, x), y)) → C(c(c(0))) at position [0] we obtained the following new rules:
C(a(a(0, y0), y1)) → C(c(0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, y0), y1)) → C(c(0))
The TRS R consists of the following rules:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, y0), y1)) → C(c(0))
The TRS R consists of the following rules:
c(y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, y0), y1)) → C(c(0))
The TRS R consists of the following rules:
c(y) → y
The set Q consists of the following terms:
c(x0)
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule C(a(a(0, y0), y1)) → C(c(0)) at position [0] we obtained the following new rules:
C(a(a(0, y0), y1)) → C(0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ Rewriting
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
C(a(a(0, y0), y1)) → C(0)
The TRS R consists of the following rules:
c(y) → y
The set Q consists of the following terms:
c(x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.