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Proof step 14 from proof valid-idx-rec (previous step)

kar[m0] = key-index(key0, m2), ar[m2] = elemindex(a, m3), m0 + 1 < # kar, m0 < # kar, m < # ar, n < # kar, m0 < n, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m, ∀ 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 ⊦ m0 = n, m2 = m, m2 < # ar ∧ m3 = m0

User applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: m0 abort: F discard: T computed: T

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ key-index(key, m).idx = m; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ elemindex(a, m).idx = m; (s) Used terms: m0