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Proof step 4 from proof hasKeys-app (
previous step
)
r -- sk ∈ st, r ∈ st, ∀ r, sks0, st. hasKeys(r, sks1 + sks0, st) ↔ hasKeys(r, sks1, st) ∧ hasKeys(r, sks0, st) ⊦ hasKeys(r, sks1 + sks0, st) ↔ hasKeys(r, sks1, st) ∧ hasKeys(r, sks0, st)
User applied rule all left
And closed the goal!
Rule argument
[EXRARG : pos: left 3, [extquanttermlist: (r, sks0, st) abort:F , discard:T , computed:T ]]
Additional information
[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, sks0, st)]]