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

⊦ r = r0, (r1 -- sk ∈ addobj(r0, class, fis, st) -- r ↔ r1 -- sk ∈ addobj(r0, class, fis, st -- r)) ∧ (addobj(r0, class, fis, st) -- r)[r1 -- sk] = (addobj(r0, class, fis, st -- r)[r1 -- sk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
⊦ r = r0, (r1 ≠ r ∧ r1 -- sk ∈ addobj(r0, class, fis, st) ↔ r1 -- sk ∈ addobj(r0, class, fis, st -- r)) ∧ (addobj(r0, class, fis, st) -- r)[r1 -- sk] = (addobj(r0, class, fis, st -- r)[r1 -- sk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st])]]