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Proof step 16 from proof addarray-addaray (previous step)

is_fskey(sk), 0 ≤ i ⊦ (r -- sk ∈ addarray(r, ty, i, st) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(r, ty, i, st)[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- sk]), sk ≠ _type ∧ sk ≠ _length

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ is_fskey(sk) ↔ (∃ fs. sk = fs ')
Proof step 15:
∃ fs. sk = fs ', 0 ≤ i ⊦ (r -- sk ∈ addarray(r, ty, i, st) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(r, ty, i, st)[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- sk]), sk ≠ _type ∧ sk ≠ _length

Rule argument

[speclemmaarg: ,is_fskey(sk) ↔ (∃ fs. sk = fs ') ,[sk] → [sk] ,F ,storekey , ,ex-03 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk)]]