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Proof step 5 from proof valid-idx-remlast (previous step)

kar[m] = key-index(key, n), m < # kar, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ m ≤ # kar[0, key-index(key, n)] ∧ (∀ m0. m0 < m → kar[0, key-index(key, n)][m0].idx < # ar[n, ar[n].idx:= 0] ∧ ar[n, ar[n].idx:= 0][kar[0, key-index(key, n)][m0].idx].idx = m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
kar[m] = key-index(key, n), m0 < # kar, m0 < m, m < # kar, m0 + 1 < # kar, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ kar[0, key-index(key, n)][m0].idx < # ar ∧ ar[n, ar[n].idx:= 0][kar[0, key-index(key, n)][m0].idx].idx = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # ar[n, ei] = # ar; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ # kar[n, ki] = # kar; (s)