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Proof step 9 from proof add-in-02 (previous step)

∀ r, r0, class1, st. r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st ⊦ r ∈ (addobj(r0, class1, fis0, st)[r0 -- sk, val]) ↔ r = r0 ∨ r ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ∈ st[r0 -- sk, val] ↔ r = r0 ∨ r ∈ st
Proof step 8:
∀ r, r0, class1, st. r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st ⊦ r = r0 ∨ r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st

Rule argument

[speclemmaarg: ,r ∈ st[r0 -- sk, val] ↔ r = r0 ∨ r ∈ st ,[val, sk, r0, st, r] → [val, sk, r0, addobj(r0, class1, fis0, st), r] ,F ,storefuns , ,instore-02 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val, sk, r0, addobj(r0, class1, fis0, st), r)]]