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Proof step 11 from proof addobj-delsw (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ sk ∈ fis → (addobj(r, class, fis, st)[r -- sk]) = fis[sk]
Proof step 9:
¬ sk ∈ fis ⊦ r = r0, (addobj(r0, class, fis, st)[r0 -- sk]) = (addobj(r0, class, fis, st -- r)[r0 -- sk])
Proof step 10:
⊦ sk ∈ fis → (addobj(r, class, fis, st)[r -- sk]) = fis[sk]

Rule argument

[crewrite-arg: add-in-09 , F , [sk, st, fis, class, r] → [sk, st, fis, class, r0] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, st, fis, class, r0)]]