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Proof step 85 from proof putarray-addarray-one (previous step)

is_reftype(ty) ⊦ r0 -- sk ∈ addarray(r, ty, n→i(# vals) + 1, st) ∨ r0 = r ∧ is_indexkey(sk) ∧ 0 < sk .index ∧ sk .index ≤ n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, refval(jvmref) ' + vals, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 62:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, i, st) ↔ r -- sk ∈ st)
Proof step 83:
¬ r0 ≠ r, is_reftype(ty) ⊦ r0 -- sk ∈ addarray(r, ty, n→i(# vals) + 1, st) ∨ r0 = r ∧ is_indexkey(sk) ∧ 0 < sk .index ∧ sk .index ≤ n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, refval(jvmref) ' + vals, st)
Proof step 84:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, vals, st) ↔ r -- sk ∈ st)

Rule argument

[crewrite-arg: in-1dim-01 , F , [st, i, ty, r0, sk, r] → [st, n→i(# vals) + 1, ty, r, sk, r0] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, n→i(# vals) + 1, ty, r, sk, r0)]]