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Proof step 9 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 ∈ st[rk, val] ∧ hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sk ' + sks0, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ hasKeys(r, sk ' + sks, st) ↔ r -- sk ∈ st ∧ hasKeys(r, sks, st)
Proof step 8:
rk ∈ st, ∀ rk, st, r, val. rk ∈ st → (hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sks0, st)) ⊦ r -- sk ∈ st[rk, val] ∧ hasKeys(r, sks0, st[rk, val]) ↔ r -- sk ∈ st ∧ hasKeys(r, sks0, st)

Rule argument

[crewrite-arg: hasKeys-rec, F, [st, sks, sk, r] → [st, sks0, sk, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, sks0, sk, r)]]