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Proof step 63 from proof addarray-addaray (previous step)

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])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ rk .ref ≠ r → (addarray(r, ty, i, st)[rk]) = st[rk]
Proof step 57:
¬ 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])
Proof step 62:
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])

Rule argument

[crewrite-arg: add-at-02, F, [rk, st, i, ty, r] → [rk, st, i, ty, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, st, i, ty, r)]]