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Proof step 14 from proof instore-02 (previous step)

⊦ (∃ sk0. r -- sk0 ∈ st[r0 -- sk, val]) ↔ r = r0 ∨ r ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 13:
⊦ (∃ sk0. r -- sk0 ∈ st[r0 -- sk, val]) ↔ r = r0 ∨ (∃ sk. r -- sk ∈ st)

Rule argument

[crewrite-arg: instore, F, [st, r] → [st, r]]

Additional information

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