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Proof step 6 from proof vals-at (previous step)

val1 ' + fis0 .vals ≠ [] ∧ n ≠ 0, ((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) .rest)[n - 1] = val

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
((sk1 × val1) ' + fis0)[n] = sk × val, n ≤ # fis0, ∀ n0, sk0, val0. fis0[n0] = sk0 × val0 ∧ ¬ # fis0 ≤ n0 → (fis0 .vals)[n0] = val0 ∨ fis0 = [] ⊦ n = 0, (fis0 .vals)[n - 1] = val

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, javavalues, e-01, ⊦ val ' + vals ≠ []], [CSIMPNAMEDSEQ : list, javavalues, a, ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPNAMEDSEQ : list, javavalues, c-07, ⊦ (val ' + vals) .rest = vals])]]