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Proof step 11 from proof valid-idx-rec (previous step)

kar[m0] = key-index(key0, m), m0 < n, n < # kar, m < # ar, m0 < # kar, m0 + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ ¬ n < m0 ∧ m1 < # kar ∧ ¬ n < m1 ∧ m0 ≠ m1 → kar[n, key-index(key, m)][m0].idx ≠ kar[n, key-index(key, m)][m1].idx, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m0 = n

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: m0, n abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ n < m → kar[m, ki][n] = kar[n]; (s)% ⊦ m < # kar → kar[m, ki][m] = ki; (s)% ⊦ key-index(key, m).idx = m; (s) Used terms: m0, n