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Proof step 12 from proof addobj-addobj-switch (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
⊦ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[r -- sk]) = (addobj(r, class, fis, st)[r -- sk]), r = r0, sk ∈ fis

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]