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Proof step 73 from proof union-addarray-rew (previous step)

newref(r0, st), newref(r0, st0) ⊦ r0 = jvmref, (r0 -- sk ∈ addarray(r0, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r0, ty, vals, st0)) ∧ (addarray(r0, ty, vals, st ∪ st0)[r0 -- sk]) = (st ∪ addarray(r0, ty, vals, st0))[r0 -- sk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ≠ r0 ∨ r = r0 ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r = r0 ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index) → (addarray(r, ty, vals, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 33:
¬ (r0 ≠ r0 ∨ r0 = r0 ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r0 = r0 ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index)), newref(r0, st), newref(r0, st0) ⊦ r0 = jvmref, (r0 -- sk ∈ addarray(r0, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r0, ty, vals, st0)) ∧ (addarray(r0, ty, vals, st ∪ st0)[r0 -- sk]) = (st ∪ addarray(r0, ty, vals, st0))[r0 -- sk]
Proof step 72:
r0 ≠ r0 ∨ r0 = r0 ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r0 = r0 ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index), newref(r0, st), newref(r0, st0) ⊦ r0 = jvmref, (r0 -- sk ∈ addarray(r0, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r0, ty, vals, st0)) ∧ (st ∪ st0)[r0 -- sk] = (st ∪ addarray(r0, ty, vals, st0))[r0 -- sk]

Rule argument

[crewrite-arg: add-at-other , F , [sk, r0, st, vals, ty, r] → [sk, r0, st ∪ st0, vals, ty, r0] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, r0, st ∪ st0, vals, ty, r0)]]