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Proof step 4 from proof keyset-subarray-eq (previous step)

subarray(kar0, 0, m0 +1) = subarray(kar1, 0, m0 +1), m0 +1 ≤ # kar1, m0 +1 ≤ # kar0, ∀ kar0, kar1. subarray(kar0, 0, m0) = subarray(kar1, 0, m0) ∧ m0 ≤ # kar0 ∧ m0 ≤ # kar1 → keyset(kar0, m0) = keyset(kar1, m0) ⊦ keyset(kar0, m0 +1) = keyset(kar1, m0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
subarray(kar0, 0, m0 + 1) = subarray(kar1, 0, m0 + 1), m0 < # kar1, m0 < # kar0, ∀ kar0, kar1. subarray(kar0, 0, m0) = subarray(kar1, 0, m0) ∧ m0 ≤ # kar0 ∧ m0 ≤ # kar1 → keyset(kar0, m0) = keyset(kar1, m0) ⊦ keyset(kar0, m0) ++ kar0[m0] = keyset(kar1, m0) ++ kar1[m0]

Rule argument

Additional information

Used simplifier rules: keyset-rec: ⊦ n < # kar → keyset(kar, n + 1) = keyset(kar, n) ++ kar[n];% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)