Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 4 ms)
↳4 AND
    ↳5 QDP
        ↳6 NonTerminationLoopProof (⇔, 13.2 s)
        ↳7 NO
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔, 0 ms)
        ↳10 YES

(0) Obligation:

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.

(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__F(X, g(X), Y) → A__F(Y, Y, Y)
MARK(f(X1, X2, X3)) → A__F(X1, X2, X3)
MARK(g(X)) → A__G(mark(X))
MARK(g(X)) → MARK(X)
MARK(b) → A__B

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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) NonTerminationLoopProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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.



(7) NO

(8) Obligation:

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.

(9) 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:

  • MARK(g(X)) → MARK(X)
    The graph contains the following edges 1 > 1

(10) YES