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Proof step 2 from proof k-dups (previous step)

dups(x) ⊦ x = [], (x.first ∈ x.rest ∨ dups(x.rest)) ∧ # x ≠ 1

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a, y] flags:

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%e-01: ⊦ ¬ dups([]);%c: ⊦ a + x = a ' + x; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%in-empty: ⊦ ¬ a ∈ []; from specification list%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a, y