YES
proof of Strategy_removed_AG01_#4.20.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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


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

   f(f(x)) -> f(x)
   g(0) -> g(f(0))

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(f(x)) -> f(x)
   0'(g(x)) -> 0'(f(g(x)))

Q is empty.

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(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:

   f(f(x)) -> f(x)
   0'(g(x)) -> 0'(f(g(x)))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
5, 6, 7, 8

Node 5 is start node and node 6 is final node.

Those nodes are connected through the following edges:

* 5 to 6 labelled f_1(0), f_1(1)* 5 to 7 labelled 0'_1(0)* 6 to 6 labelled #_1(0)* 7 to 8 labelled f_1(0)* 8 to 6 labelled g_1(0)


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(4)
YES