YES
proof of /TPDB/Der95_03.ari
# AProVE Commit ID: 23a904c96b029b0a549cde0d0d17dbccf967db59 jckassing 20260626 unpublished dirty


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

(0) QTRS
(1) Strip Symbols Proof [SOUND, 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)) -> g(f(x))

Q is empty.

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(1) Strip Symbols Proof (SOUND)
We were given the following TRS:

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

By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: 

   f(x) -> g(x)

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

   f(x) -> 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 0. This implies Q-termination of R.
The following rules were used to construct the certificate:

   f(x) -> g(x)

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
1, 4

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

Those nodes are connected through the following edges:

* 1 to 4 labelled g_1(0)* 4 to 4 labelled #_1(0)


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(4)
YES