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Proof step 7 from proof keys-at (previous step)

sk1 ' + fis0 .keys ≠ [] ∧ n ≠ 0, ((sk1 × val1) ' + fis0)[n] = sk × val, n ≤ # fis0, ∀ n0, sk0, val0. fis0[n0] = sk0 × val0 ∧ ¬ # fis0 ≤ n0 → (fis0 .keys)[n0] = sk0 ∨ fis0 = [] ⊦ ((sk1 ' + fis0 .keys) .rest)[n - 1] = sk

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((sk1 × val1) ' + fis0)[n] = sk × val, n ≤ # fis0, ∀ n0, sk0, val0. fis0[n0] = sk0 × val0 ∧ ¬ # fis0 ≤ n0 → (fis0 .keys)[n0] = sk0 ∨ fis0 = [] ⊦ n = 0, (fis0 .keys)[n - 1] = sk

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, e-01, ⊦ sk ' + sks ≠ []], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, c-07, ⊦ (sk ' + sks) .rest = sks])]]