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Proof step 217 from proof putarray-addarray-override (previous step)

i + n→i(# vals) ≤ n→i(# vals0), 0 ≤ i ⊦ ∀ rk. (rk ∈ putArray(vals, r, i, addarray(r, ty, vals0, st)) ↔ rk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)) ∧ (putArray(vals, r, i, addarray(r, ty, vals0, st))[rk]) = (addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)[rk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 216:
0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ (rk ∈ putArray(vals, r, i, addarray(r, ty, vals0, st)) ↔ rk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)) ∧ (putArray(vals, r, i, addarray(r, ty, vals0, st))[rk]) = (addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)[rk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]