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Proof step 24 from proof addobj-delsw (previous step)

r1 ≠ r0 ⊦ r = r0, (r1 ≠ r ∧ r1 -- sk ∈ addobj(r0, class, fis, st) ↔ r1 -- sk ∈ 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 13:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 14:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 23:
⊦ r = r0, r1 = r0, (addobj(r0, class, fis, st) -- r)[r1 -- sk] = (st -- r)[r1 -- sk]

Rule argument

[emptyarg]

Additional information

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