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Proof step 8 from proof no-dups-idx-remlast (previous step)

kar[m0] = key-index(key, m2), kar[m] = key-index(key0, m2), m < # kar, m0 < # kar, m0 < m, m0 + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ ¬ m < m0 ∧ m1 < # kar ∧ ¬ m < m1 ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ # kar = 1, m0 = 0

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ key-index(key, m).idx = m; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s) Used terms: m, m0