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Proof step 23 from proof addobj-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 16:
¬ r0 ≠ r, newref(r, st0), newref(r, st) ⊦ r = jvmref, (r0 -- sk ∈ addobj(r, class, fis, st) ∪ st0 ↔ r0 -- sk ∈ addobj(r, class, fis, st ∪ st0)) ∧ (addobj(r, class, fis, st) ∪ st0)[r0 -- sk] = (addobj(r, class, fis, st ∪ st0)[r0 -- sk])
Proof step 22:
r0 ≠ r, newref(r, st0), newref(r, st) ⊦ r = jvmref, (r0 -- sk ∈ addobj(r, class, fis, st) ∪ st0 ↔ r0 -- sk ∈ st ∪ st0) ∧ (addobj(r, class, fis, st) ∪ st0)[r0 -- sk] = (addobj(r, class, fis, st ∪ st0)[r0 -- sk])

Rule argument

[crewrite-arg: add-in-04 , F , [st, fis, class1, r0, sk, r] → [st ∪ st0, fis, class, r, sk, r0] ]

Additional information

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