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[moca] confused by Sangiogi & Walker's ``up to'' lemmas
Dear pi friends (and Davide and David in particular!),
I've been reading the Sangiorgi and Walker pi book, and became
confused by their presentation of ``up to'' techniques. There appears
to be (an apparently) superfluous conditions in a key definition. I
may well be missing something and would be grateful for any comments.
They first say (Def 2.3.1, p.82) that a relation R _strongly
progresses_ to S, written R~>S if for all (P,Q) in R and P-{alpha}->P'
there exists Q' such that Q-{alpha}->Q' and (P',Q') in S:
alpha
P ------------> P'
| .
|R .S
| alpha .
Q ............> Q'
Then go on to define that a function F on relations is _strongly safe_
if
R subseteq S and R~>S implies F(R) subseteq F(S) and F(R)~>F(S)
My problem: I don't see the use of the monotonicity with respect to
subseteq. In particular, Lemma 2.3.8,
If F is strongly safe and R~>F(R) then R and F(R) are included in ~
(strong bisimilarity).
seems to have a simpler proof (requiring no use of
subseteq-monotonicity) if recast as follows:
If F is monotonic wrt ~> and R~>F(R) then R and F(R) are included
in ~.
My proof:
Let F^{omega}(R) = Union_{n in omega} F^n(R).
Claim: F^{omega}(R) is a bisimulation.
Proof of claim: Suppose (P,Q) in F^{omega}(R) and P-{alpha}->P'.
By definition, there exists i in omega such that (P,Q) in
F^i(R). Since F is monotonic wrt ~> and R~>F(R), we have
F^i(R)~>F^{i+1}(R) by repeated application of F. Thus, there
exists Q' such that Q-{alpha}->Q' and (P',Q') in F^{i+1}(R).
Hence (P',Q') in F^{omega}(R).
End of Claim.
Thus, R union F(R) subseteq F^{omega}(R) subseteq ~, as desired.
End of proof.
Comments?
Warm regards,
-James Leifer
INRIA Rocquencourt
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