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Proof step 10 from proof le (previous step)

a ∈ y, # y ≤ # filter(rmdup(x0), y), #(y -1l a) < # filter(rmdup(x0), y -1l a) → filter(rmdup(x0), y -1l a) = [] ⊦ a ∈ x0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del1-01: ⊦ ¬ a ∈ x → filter(rmdup(x), y -1l a) = filter(rmdup(x), y);%filter-empty-01: ⊦ filter(x, []) = [];% ⊦ ¬ a ∈ []; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ # [] = 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ [] -1l a = []; (s)% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ n -1 = n - 1; (s)% ⊦ a ∈ x → #(x -1l a) = (# x)-1; (s)