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Proof step 5 from proof update-atin (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
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]

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

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