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Proof step 6 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) ↔ hasKeys(r, sk ' + sks1, st) ∧ hasKeys(r, sks0, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ hasKeys(r, sk ' + sks, st) ↔ r -- sk ∈ st ∧ hasKeys(r, sks, st)
Proof step 5:
∀ 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)

Rule argument

[crewrite-arg: hasKeys-rec, F, [st, sks, sk, r] → [st, sks1, sk, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, sks1, sk, r)]]