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Proof step 6 from proof l-02 (previous step)

# x + # y0 = # y0 - 1, a ∈ y0, ∀ x, a. # x + # y0 = #(y0 -1l a) → x = [] ∧ ¬ a ∈ y0 ⊦ a = a0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# y0, m]

Additional information

Used simplifier rules: ⊦ m = m + n ↔ n = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ # [] = 0; (s)% ⊦ n + 0 = n; (s)% ⊦ ¬ a ∈ []; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # y0, m