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Proof step 18 from proof no-dups-idx-rec (previous step)

kar[n, key-index(key, m2)][m] = key-index(key1, m3), kar[n, key-index(key, m2)][m0] = key-index(key0, m3), m = n, n < # kar, m < # kar, m0 < # kar, m ≤ n, m0 ≤ n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m0 = m, ∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
kar[m0] = key-index(key0, m3), m0 < n, n < # kar, m0 < # kar, m0 + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m0 = n, ∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m3

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ m; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ key-index(key, m) = key-index(key0, m0) ↔ key = key0 ∧ m = m0; (ws)% ⊦ m < # kar → kar[m, ki][m] = ki; (s)% ⊦ m ≠ n → kar[n, ki][m] = kar[m]; (s)