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

⊦ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[r -- sk]) = (addobj(r, class, fis, st)[r -- sk]), r = r0, sk ∈ fis

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ¬ sk ∈ fis ∧ sk ≠ _type → (addobj(r, class, fis, st)[r -- sk]) = st[r -- sk]
Proof step 8:
¬ _type ∈ fis ⊦ (addobj(r, class1, fis, st)[r -- _type]) = typeval(mkclasstype(class1))
Proof step 9:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 10:
⊦ ¬ sk ∈ fis ∧ sk ≠ _type → (addobj(r, class, fis, st)[r -- sk]) = st[r -- sk]

Rule argument

[crewrite-arg: add-notin , F , [sk, st, fis, class, r] → [sk, addobj(r0, class0, fis0, st), fis, class, r] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ typeval(ty) = typeval(ty0) ↔ ty = ty0])], [EXTRATERMS : (sk, addobj(r0, class0, fis0, st), fis, class, r)]]