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Proof step 6 from proof add-in-09 (previous step)

fv = sk0 × val, sk ∈ fv ' + fis0, ∀ sk, r, class, st. sk ∈ fis0 → (addobj(r, class, fis0, st)[r -- sk]) = fis0[sk] ⊦ ((sk = sk0 ⊃ val; addobj(r, class, fis0, st)[r -- sk])) = (fv ' + fis0)[sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
sk ∈ fis0, ∀ sk, r, class, st. sk ∈ fis0 → (addobj(r, class, fis0, st)[r -- sk]) = fis0[sk] ⊦ ((sk = sk0 ⊃ val; addobj(r, class, fis0, st)[r -- sk])) = ((sk0 × val) ' + fis0)[sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : sk ≠ sk0 ⊦ sk ∈ (sk0 × val) ' + fis ↔ sk ∈ fis], [CSIMPSEQ : ⊦ ((sk × val) ' + fis)[sk] = val], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]