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Proof step 4 from proof hasKeys-all (previous step)

r -- sk ∈ st, hasKeys(r, sks0, st), sk0 ∈ sks0, r ∈ st, hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st), ∀ r, st. hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st) ⊦ r -- sk0 ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
r -- sk ∈ st, hasKeys(r, sks0, st), sk0 ∈ sks0, r ∈ st, ∀ r, st. hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st), ∀ sk. sk ∈ sks0 → r -- sk ∈ st ⊦ r -- sk0 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]