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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPSizeChangeProof (⇔)
↳4 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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(g(X), Y) → A__F(mark(X), f(g(X), Y))
A__F(g(X), Y) → MARK(X)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:


POL(a__f(x1, x2)) = 1 + x1   
POL(f(x1, x2)) = 1 + x1   
POL(g(x1)) = 1 + x1   
POL(mark(x1)) = x1   

From the DPs we obtained the following set of size-change graphs:

  • A__F(g(X), Y) → A__F(mark(X), f(g(X), Y)) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1

  • A__F(g(X), Y) → MARK(X) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MARK(f(X1, X2)) → A__F(mark(X1), X2) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1

  • MARK(f(X1, X2)) → MARK(X1) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MARK(g(X)) → MARK(X) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05].


mark(g(X)) → g(mark(X))
mark(f(X1, X2)) → a__f(mark(X1), X2)
a__f(X1, X2) → f(X1, X2)
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))

(4) TRUE