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Proof step 6 from proof update-atin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ update(r, (sk × val) ' + fis, st) = (update(r, fis, st)[r -- sk, val])
Proof step 5:
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]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : sk ≠ sk0 ⊦ sk ∈ (sk0 × val) ' + fis ↔ sk ∈ fis], [CSIMPSEQ : ⊦ ((sk × val) ' + fis)[sk] = val], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]