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

¬ r1 ≠ r0 ⊦ 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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ addobj(r, class, fis, st) ↔ ¬ ¬ (sk ∈ fis ∨ sk = _type ∨ r -- sk ∈ st)
Proof step 11:
⊦ r = r0, (addobj(r0, class, fis, st)[r0 -- sk]) = (addobj(r0, class, fis, st -- r)[r0 -- sk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ≠ r0 → (st -- r)[r0 -- sk0] = st[r0 -- sk0]], [CSIMPSEQ : ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st])]]