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Proof step 62 from proof addarray-addaray (previous step)

rk .ref ≠ r, 0 ≤ i ⊦ (rk ∈ addarray(r, ty, i, st) ↔ rk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ st[rk] = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[rk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
⊦ rk .ref ≠ r → (addarray(r, ty, vals, st)[rk]) = st[rk]
Proof step 61:
0 ≤ i ⊦ rk .ref = r, rk ∈ addarray(r, ty, i, st) ↔ rk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]