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Proof step 11 from proof update-keyin (conclusion)
⊦ ¬ sk ∈ fis → (r0 -- sk ∈ update(r, fis, st) ↔ r0 -- sk ∈ st)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 10
:
⊦ r0 -- sk ∈ update(r, fis, st) ↔ r0 -- sk ∈ st, sk ∈ fis
Rule argument
[emptyarg]
Additional information
[[CONCRETESIMPRULES : ()]]