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Proof step 9 from proof addobj-delsw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ (addobj(r0, class, fis, st)[r0 -- sk]) = (addobj(r0, class, fis, st -- r)[r0 -- sk]), r = r0, sk ∈ fis

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]