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Proof step 48 from proof putarray-add-same (previous step)

⊦ (r0 -- sk ∈ putArray(vals, r, addarray(r, ty, n→i(# vals), st)) ↔ r0 -- sk ∈ addarray(r, ty, vals, st)) ∧ (putArray(vals, r, addarray(r, ty, n→i(# vals), st))[r0 -- sk]) = (addarray(r, ty, vals, st)[r0 -- sk])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ putArray(vals, r, st) = putArray(vals, r, 0, st)
Proof step 47:
⊦ (r0 -- sk ∈ putArray(vals, r, 0, addarray(r, ty, n→i(# vals), st)) ↔ r0 -- sk ∈ addarray(r, ty, vals, st)) ∧ (putArray(vals, r, 0, addarray(r, ty, n→i(# vals), st))[r0 -- sk]) = (addarray(r, ty, vals, st)[r0 -- sk])

Rule argument

[speclemmaarg: ,putArray(vals, r, st) = putArray(vals, r, 0, st) ,[st, r, vals] → [addarray(r, ty, n→i(# vals), st), r, vals] ,F ,storearrays , ,putArray-all ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarray(r, ty, n→i(# vals), st), r, vals)]]