Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b
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:
MARK(b) → A__B
MARK(g(X)) → A__G(mark(X))
MARK(f(X1, X2, X3)) → A__F(X1, X2, X3)
A__F(X, g(X), Y) → A__F(Y, Y, Y)
MARK(g(X)) → MARK(X)
The TRS R consists of the following rules:
a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b
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:
MARK(b) → A__B
MARK(g(X)) → A__G(mark(X))
MARK(f(X1, X2, X3)) → A__F(X1, X2, X3)
A__F(X, g(X), Y) → A__F(Y, Y, Y)
MARK(g(X)) → MARK(X)
The TRS R consists of the following rules:
a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ NonTerminationProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A__F(X, g(X), Y) → A__F(Y, Y, Y)
The TRS R consists of the following rules:
a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
A__F(X, g(X), Y) → A__F(Y, Y, Y)
The TRS R consists of the following rules:
a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b
s = A__F(a__g(a__b), a__g(a__b), Y) evaluates to t =A__F(Y, Y, Y)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [Y / a__g(a__b)]
Rewriting sequence
A__F(a__g(a__b), a__g(a__b), a__g(a__b)) → A__F(a__g(a__b), a__g(c), a__g(a__b))
with rule a__b → c at position [1,0] and matcher [ ]
A__F(a__g(a__b), a__g(c), a__g(a__b)) → A__F(a__g(a__b), g(c), a__g(a__b))
with rule a__g(X) → g(X) at position [1] and matcher [X / c]
A__F(a__g(a__b), g(c), a__g(a__b)) → A__F(a__g(b), g(c), a__g(a__b))
with rule a__b → b at position [0,0] and matcher [ ]
A__F(a__g(b), g(c), a__g(a__b)) → A__F(c, g(c), a__g(a__b))
with rule a__g(b) → c at position [0] and matcher [ ]
A__F(c, g(c), a__g(a__b)) → A__F(a__g(a__b), a__g(a__b), a__g(a__b))
with rule A__F(X, g(X), Y) → A__F(Y, Y, Y)
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
MARK(g(X)) → MARK(X)
The TRS R consists of the following rules:
a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b
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
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
MARK(g(X)) → MARK(X)
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:
- MARK(g(X)) → MARK(X)
The graph contains the following edges 1 > 1