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Proof step 47 from proof putarray-add-same (previous step)

⊦ (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])

User applied rule case distinction

And obtained the following new goals

Proof step 21:
⊦ r0 -- sk ∈ putArray(vals, r, 0, addarray(r, ty, n→i(# vals), st)) ↔ r0 -- sk ∈ addarray(r, ty, vals, st)
Proof step 46:
⊦ (putArray(vals, r, 0, addarray(r, ty, n→i(# vals), st))[r0 -- sk]) = (addarray(r, ty, vals, st)[r0 -- sk])

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]