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Proof step 33 from proof addobj-addobj-switch (previous step)

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 32:
⊦ r = r0, r1 = r, (r1 -- sk ∈ addobj(r0, class0, fis0, st) ↔ r1 -- sk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r0, class0, fis0, st)[r1 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r1 -- sk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]