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Proof step 7 from proof hasKeys-put-01 (previous step)

rk ∈ st, ∀ rk, st, r, val. rk ∈ st → (hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sks0, st)) ⊦ (r -- sk = rk ∨ r -- sk ∈ st) ∧ hasKeys(r, sks0, st[rk, val]) ↔ r -- sk ∈ st ∧ hasKeys(r, sks0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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)

Rule argument

[emptyarg]

Additional information

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