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Proof step 6 from proof hasKeys-put-02 (previous step)

∀ r, st, sk, val. (hasKeys(r, sks0, st[r -- sk, val]) ↔ hasKeys(r, sks0, st)) ∨ sk ∈ sks0 ⊦ r -- sk0 ∈ st ∧ hasKeys(r, sks0, st[r -- sk, val]) ↔ hasKeys(r, sk0 ' + sks0, st), sk = sk0, sk ∈ sks0

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 5:
∀ r, st, sk, val. (hasKeys(r, sks0, st[r -- sk, val]) ↔ hasKeys(r, sks0, st)) ∨ sk ∈ sks0 ⊦ r -- sk0 ∈ st ∧ hasKeys(r, sks0, st[r -- sk, val]) ↔ r -- sk0 ∈ st ∧ hasKeys(r, sks0, st), sk = sk0, sk ∈ sks0

Rule argument

[crewrite-arg: hasKeys-rec, F, [st, sks, sk, r] → [st, sks0, sk0, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, sks0, sk0, r)]]