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Proof step 57 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)) ∧ (addarray(r, ty, i, 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 56:
0 ≤ i ⊦ (rk ∈ addarray(rk .ref, ty, i, st) ↔ rk ∈ addarray(rk .ref, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(rk .ref, ty, i, st)[rk]) = (addarray(rk .ref, ty, mklist(initial_value(ty), i→n(i)), st)[rk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]