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Proof step 10 from proof addobj-at-other-03 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ sk = _type, (addobj(r, class, fis, st)[r0 -- sk]) = st[r0 -- sk], sk ∈ fis

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]