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Proof step 19 from proof add-dupput (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ sk ∈ fis → (addobj(r, class, fis, st)[r -- sk]) = fis[sk]
Proof step 18:
sk0 ∈ fis + fis0, sk0 ∈ fis + (sk × val) ' + fis0, sk ∈ fis ⊦ (fis + (sk × val) ' + fis0)[sk0] = (fis + fis0)[sk0]

Rule argument

[crewrite-arg: add-in-09 , 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 ∈ fis → (fis + fis0)[sk] = fis0[sk]], [CSIMPSEQ : ⊦ sk ∈ (sk0 × val) ' ↔ sk = sk0], [CSIMPSEQ : ⊦ ((sk × val) ' + fis)[sk] = val])], [EXTRATERMS : (sk0, st, fis + fis0, class1, r0)]]