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Proof step 34 from proof addobj-addobj-switch (previous step)

⊦ r = r0, (r1 -- sk ∈ addobj(r, class, fis, addobj(r0, class0, fis0, st)) ↔ r1 -- sk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[r1 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r1 -- 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 15:
¬ r1 ≠ r ⊦ r = r0, (r1 -- sk ∈ addobj(r, class, fis, addobj(r0, class0, fis0, st)) ↔ r1 -- sk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[r1 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r1 -- sk])
Proof step 33:
r1 ≠ r ⊦ r = r0, (r1 -- sk ∈ addobj(r0, class0, fis0, st) ↔ r1 -- sk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[r1 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r1 -- sk])

Rule argument

[crewrite-arg: add-in-04 , F , [st, fis, class1, r0, sk, r] → [addobj(r0, class0, fis0, st), fis, class, r, sk, r1] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addobj(r0, class0, fis0, st), fis, class, r, sk, r1)]]