YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex15_Luc06_Z.ari

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 RFCMatchBoundsTRSProof (⇔, 0 ms)
↳4 YES

(0) Obligation:

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

f(n__f(n__a)) → f(n__g(f(n__a)))
f(X) → n__f(X)
a → n__a
g(X) → n__g(X)
activate(n__f(X)) → f(X)
activate(n__a) → a
activate(n__g(X)) → g(X)
activate(X) → X

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

n__a'(n__f(f(x))) → n__a'(f(n__g(f(x))))
f(X) → n__f(X)
a'(x) → n__a'(x)
g(X) → n__g(X)
n__f(activate(X)) → f(X)
n__a'(activate(x)) → a'(x)
n__g(activate(X)) → g(X)
activate(X) → X

Q is empty.

(3) RFCMatchBoundsTRSProof (EQUIVALENT transformation)

Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2. This implies Q-termination of R.
The following rules were used to construct the certificate:

n__a'(n__f(f(x))) → n__a'(f(n__g(f(x))))
f(X) → n__f(X)
a'(x) → n__a'(x)
g(X) → n__g(X)
n__f(activate(X)) → f(X)
n__a'(activate(x)) → a'(x)
n__g(activate(X)) → g(X)
activate(X) → X

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

1, 4, 8, 9, 10, 14, 15, 16

Node 1 is start node and node 4 is final node.

Those nodes are connected through the following edges:

  • 1 to 8 labelled n__a'_1(0)
  • 1 to 4 labelled n__f_1(0), n__a'_1(0), n__g_1(0), f_1(0), a'_1(0), g_1(0), activate_1(0), n__f_1(1), n__a'_1(1), n__g_1(1), f_1(1), a'_1(1), g_1(1), activate_1(1), n__f_1(2), n__a'_1(2), n__g_1(2)
  • 1 to 14 labelled n__a'_1(1)
  • 4 to 4 labelled #_1(0)
  • 8 to 9 labelled f_1(0), n__f_1(1)
  • 9 to 10 labelled n__g_1(0)
  • 10 to 4 labelled f_1(0), n__f_1(1), f_1(1), n__f_1(2)
  • 14 to 15 labelled f_1(1), n__f_1(2)
  • 15 to 16 labelled n__g_1(1)
  • 16 to 4 labelled f_1(1), n__f_1(2)

(4) YES