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Proof step 5 from proof hasKeys-app (previous step)

∀ r, sks0, st. hasKeys(r, sks1 + sks0, st) ↔ hasKeys(r, sks1, st) ∧ hasKeys(r, sks0, st) ⊦ r -- sk ∈ st ∧ hasKeys(r, sks1 + sks0, st) ↔ r -- sk ∈ st ∧ hasKeys(r, sks1, st) ∧ hasKeys(r, sks0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

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