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Proof step 65 from proof addarray-addaray (previous step)

0 ≤ i ⊦ addarray(r, ty, i, st) = addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 64:
0 ≤ i ⊦ ∀ rk. (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])

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addarray(r, ty, mklist(initial_value(ty), i→n(i)), st), addarray(r, ty, i, st)] ,F ,store ,javastore-ostorea ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st), addarray(r, ty, i, st))]]