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Proof step 10 from proof add-dupput (previous step)

sk ∈ fis ⊦ (addobj(r0, class1, fis + (sk × val) ' + fis0, st)[r0 -- sk0]) = (addobj(r0, class1, fis + fis0, st)[r0 -- sk0]), sk0 ∈ fis0, sk0 ∈ fis

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ¬ sk ∈ fis ∧ sk ≠ _type → (addobj(r, class, fis, st)[r -- sk]) = st[r -- sk]
Proof step 6:
¬ _type ∈ fis ⊦ (addobj(r, class1, fis, st)[r -- _type]) = typeval(mkclasstype(class1))
Proof step 9:
¬ sk0 ∈ fis + fis0 ∧ sk0 ≠ _type, sk ∈ fis ⊦ (addobj(r0, class1, fis + (sk × val) ' + fis0, st)[r0 -- sk0]) = st[r0 -- sk0], sk0 ∈ fis0, sk0 ∈ fis

Rule argument

[crewrite-arg: add-notin , F , [sk, st, fis, class, r] → [sk0, st, fis + fis0, class1, r0] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ∈ fis + fis0 ↔ sk ∈ fis ∨ sk ∈ fis0], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ sk ∈ (sk0 × val) ' ↔ sk = sk0], [CSIMPSEQ : ⊦ typeval(ty) = typeval(ty0) ↔ ty = ty0])], [EXTRATERMS : (sk0, st, fis + fis0, class1, r0)]]