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Proof step 8 from proof hasKeys-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
r -- sk ∈ st, sk0 ∈ sks0, r ∈ st, ∀ sk0. sk0 ∈ sks0 → r -- sk0 ∈ st, ∀ r, st. hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st) ⊦ hasKeys(r, sks0, st), r -- sk0 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]