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

Additional information

Used simplifier rules: eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 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%size-01: ⊦ # [] = 0; from specification list-data%add-base: ⊦ n + 0 = n; from specification nat-basic2%in-empty: ⊦ ¬ a ∈ []; from specification list%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # y0, m