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Proof step 7 from proof in-1dim-09 (previous step)

r ∈ st ⊦ ∃ sk. r -- sk ∈ addarray(r0, val, i, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 6:
∃ sk. r -- sk ∈ st ⊦ ∃ sk. r -- sk ∈ addarray(r0, val, i, st)

Rule argument

[speclemmaarg: ,r ∈ st ↔ (∃ sk. r -- sk ∈ st) ,[st, r] → [st, r] ,F ,storefuns , ,instore ]

Additional information

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