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Proof step 6 from proof add-in-08 (previous step)

∀ r, sk, class, st. r -- sk ∈ addobj(r, class, fis0, st) ↔ sk ∈ fis0 ∨ sk = _type ∨ r -- sk ∈ st ⊦ r -- sk0 ∈ (addobj(r, class, fis0, st)[r -- sk, val]) ↔ sk0 ∈ fis0 ∨ sk0 = _type ∨ r -- sk0 ∈ st ∨ sk0 = sk

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
sk ≠ sk0 ⊦ r -- sk ∈ st[r -- sk0, val] ↔ r -- sk ∈ st
Proof step 5:
sk0 ≠ sk, ∀ r, sk, class, st. r -- sk ∈ addobj(r, class, fis0, st) ↔ sk ∈ fis0 ∨ sk = _type ∨ r -- sk ∈ st ⊦ r -- sk0 ∈ addobj(r, class, fis0, st) ↔ sk0 ∈ fis0 ∨ sk0 = _type ∨ r -- sk0 ∈ st ∨ sk0 = sk

Rule argument

[speclemmaarg: sk ≠ sk0 ,r -- sk ∈ st[r -- sk0, val] ↔ r -- sk ∈ st ,[val, sk0, st, sk, r] → [val, sk, addobj(r, class, fis0, st), sk0, r] ,F ,javastore , ,instore-02 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : sk ≠ sk0 ⊦ r -- sk ∈ st[r -- sk0, val] ↔ r -- sk ∈ st], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st])], [EXTRATERMS : (val, sk, addobj(r, class, fis0, st), sk0, r)]]