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Proof step 32 from proof addobj-union-rew (previous step)

newref(r, st), newref(r, st0) ⊦ r = jvmref, r = r0, (st ∪ st0)[r0 -- sk] = (st ∪ addobj(r, class, fis, st0))[r0 -- sk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ ¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk]
Proof step 29:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 30:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 31:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)

Rule argument

[speclemmaarg: ,¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk] ,[rk, st0, st] → [r0 -- sk, addobj(r, class, fis, st0), st] ,F ,storefunsx , ,union-at-left ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]], [CSIMPSEQ : ⊦ ¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk]])], [EXTRATERMS : (r0 -- sk, addobj(r, class, fis, st0), st)]]