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Proof step 58 from proof putarray-addarray-override (previous step)

0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0), ∃ fs. sk = fs ' ⊦ r0 -- sk ∈ addarray(r, ty, vals0, st) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ r0 -- fs ' ∈ addarray(r, ty, vals0, st) ↔ r0 -- fs ' ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]