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Proof step 9 from proof le (previous step)

a ∈ y, # y ≤ # filter(rmdup(x0), y), ∀ y. # y < # filter(rmdup(x0), y) → filter(rmdup(x0), y) = [] ⊦ a ∈ x0

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: y -1l a abort: F discard: T computed: F

Additional information

Used simplifier rules: del1-01: ⊦ ¬ a ∈ x → filter(rmdup(x), y -1l a) = filter(rmdup(x), y);%filter-empty-01: ⊦ filter(x, []) = [];%l: ⊦ a ∈ x → #(x -1l a) = (# x)-1; from specification list-del%pred-03: ⊦ n -1 = n - 1; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%l: ⊦ # x = 0 ↔ x = []; from specification list-data%del1-e: ⊦ [] -1l a = []; from specification list-del%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%size-01: ⊦ # [] = 0; from specification list-data%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%in-empty: ⊦ ¬ a ∈ []; from specification list Used terms: y -1l a