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Proof step 20 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])

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 11:
¬ sk0 ∈ fis + (sk × val) ' + fis0, sk ∈ fis ⊦ (addobj(r0, class1, fis + (sk × val) ' + fis0, st)[r0 -- sk0]) = (addobj(r0, class1, fis + fis0, st)[r0 -- sk0])
Proof step 19:
sk0 ∈ fis + (sk × val) ' + fis0, sk ∈ fis ⊦ (fis + (sk × val) ' + fis0)[sk0] = (addobj(r0, class1, fis + fis0, st)[r0 -- sk0])

Rule argument

[crewrite-arg: add-in-09 , F , [sk, st, fis, class, r] → [sk0, st, fis + (sk × val) ' + fis0, class1, r0] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk0, st, fis + (sk × val) ' + fis0, class1, r0)]]