YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms]
(4) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(g(X)) -> g(f(f(X)))
   f(h(X)) -> h(g(X))

Q is empty.

----------------------------------------

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(f(X)) -> f(f(g(X)))
   h(f(X)) -> g(h(X))

Q is empty.

----------------------------------------

(3) RFCMatchBoundsTRSProof (EQUIVALENT)
Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R.
The following rules were used to construct the certificate:

   g(f(X)) -> f(f(g(X)))
   h(f(X)) -> g(h(X))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
57, 58, 59, 60, 61, 62, 63, 64

Node 57 is start node and node 58 is final node.

Those nodes are connected through the following edges:

* 57 to 59 labelled f_1(0)* 57 to 61 labelled g_1(0)* 58 to 58 labelled #_1(0)* 59 to 60 labelled f_1(0)* 60 to 58 labelled g_1(0)* 60 to 62 labelled f_1(1)* 61 to 58 labelled h_1(0)* 61 to 64 labelled g_1(1)* 62 to 63 labelled f_1(1)* 63 to 58 labelled g_1(1)* 63 to 62 labelled f_1(1)* 64 to 58 labelled h_1(1)* 64 to 64 labelled g_1(1)


----------------------------------------

(4)
YES