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Proof step 6 from proof hasKeys-put-01 (
previous step
)
rk ∈ st, r -- sk ∈ st, r ∈ st, ∀ rk, st, r, val. rk ∈ st → (hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sks0, st)) ⊦ hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sks0, st)
User applied rule all left
And closed the goal!
Rule argument
[EXRARG : pos: left 4, [extquanttermlist: (rk, st, r, val) abort:F , discard:T , computed:T ]]
Additional information
[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, st, r, val)]]