YES
proof of AG01_#3.7.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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:

   half(0) -> 0
   half(s(s(x))) -> s(half(x))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(half(x))))

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:

   0'(half(x)) -> 0'(x)
   s(s(half(x))) -> half(s(x))
   0'(s(log(x))) -> 0'(x)
   s(s(log(x))) -> half(s(log(s(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:

   0'(half(x)) -> 0'(x)
   s(s(half(x))) -> half(s(x))
   0'(s(log(x))) -> 0'(x)
   s(s(log(x))) -> half(s(log(s(x))))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
3, 4, 5, 8, 11, 12, 14, 18, 19, 20

Node 3 is start node and node 4 is final node.

Those nodes are connected through the following edges:

* 3 to 4 labelled 0'_1(0), 0'_1(1)* 3 to 5 labelled half_1(0)* 3 to 8 labelled half_1(0)* 4 to 4 labelled #_1(0)* 5 to 4 labelled s_1(0)* 5 to 14 labelled half_1(1)* 5 to 18 labelled half_1(1)* 8 to 11 labelled s_1(0)* 11 to 12 labelled log_1(0)* 12 to 4 labelled s_1(0)* 12 to 14 labelled half_1(1)* 12 to 18 labelled half_1(1)* 14 to 4 labelled s_1(1)* 14 to 14 labelled half_1(1)* 14 to 18 labelled half_1(1)* 18 to 19 labelled s_1(1)* 19 to 20 labelled log_1(1)* 20 to 4 labelled s_1(1)* 20 to 14 labelled half_1(1)* 20 to 18 labelled half_1(1)


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(4)
YES