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Proof step 34 from proof addobj-union-rew (previous step)

newref(r, st), newref(r, st0) ⊦ r = jvmref, (r0 -- sk ∈ addobj(r, class, fis, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addobj(r, class, fis, st0)) ∧ (addobj(r, class, fis, 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 2:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 26:
¬ r ≠ r0, newref(r, st), newref(r, st0) ⊦ r = jvmref, (r0 -- sk ∈ addobj(r, class, fis, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addobj(r, class, fis, st0)) ∧ (addobj(r, class, fis, st ∪ st0)[r0 -- sk]) = (st ∪ addobj(r, class, fis, st0))[r0 -- sk]
Proof step 33:
r ≠ r0, newref(r, st), newref(r, st0) ⊦ r = jvmref, (r0 -- sk ∈ addobj(r, class, fis, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addobj(r, class, fis, st0)) ∧ (st ∪ st0)[r0 -- sk] = (st ∪ addobj(r, class, fis, st0))[r0 -- sk]

Rule argument

[crewrite-arg: add-ref , F , [sk, r0, st, fis, class1, r] → [sk, r0, st ∪ st0, fis, class, r] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, r0, st ∪ st0, fis, class, r)]]