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Proof step 26 from proof addobj-addobj-switch (previous step)

¬ sk ∈ fis0 ⊦ r0 = r, (addobj(r0, class0, fis0, st)[r0 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r0 -- sk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
⊦ (addobj(r0, class0, fis0, st)[r0 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r0 -- sk]), r0 = r, sk ∈ fis0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]