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Proof step 8 from proof vals-at (previous step)

fv = sk1 × val1, (fv ' + fis0)[n] = sk × val, n ≤ # fis0, ∀ n0, sk0, val0. fis0[n0] = sk0 × val0 ∧ ¬ # fis0 ≤ n0 → (fis0 .vals)[n0] = val0 ∨ fis0 = [] ⊦ (val1 ' + fis0 .vals)[n] = val

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
((sk1 × val1) ' + fis0)[n] = sk × val, n ≤ # fis0, ∀ n0, sk0, val0. fis0[n0] = sk0 × val0 ∧ ¬ # fis0 ≤ n0 → (fis0 .vals)[n0] = val0 ∨ fis0 = [] ⊦ (val1 ' + fis0 .vals)[n] = val

Rule argument

[emptyarg]

Additional information

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