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Proof step 4 from proof update-atin (previous step)

sk ≠ sk0, sk ∈ fis0, ∀ sk, r, st. sk ∈ fis0 → (update(r, fis0, st)[r -- sk]) = fis0[sk] ⊦ (update(r, fis0, st)[r -- sk]) = ((sk0 × val) ' + fis0)[sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
sk ∈ fis0, ∀ sk, r, st. sk ∈ fis0 → (update(r, fis0, st)[r -- sk]) = fis0[sk] ⊦ sk = sk0, (update(r, fis0, st)[r -- sk]) = fis0[sk]

Rule argument

[emptyarg]

Additional information

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