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Proof step 21 from proof nodups-butlastn (previous step)

a ∈ butlastn(n, a0 ' + x), n ≤ 1 + # x, ∀ a, n. a ∈ butlastn(n, x) ∧ n ≤ # x → a ∈ x ⊦ a = a0, a ∈ x, n ≤ # x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
a ∈ butlastn(# x, (a0 ' + x).butlast), ∀ a, n. a ∈ butlastn(n, x) ∧ n ≤ # x → a ∈ x ⊦ a = a0, a ∈ x

Rule argument

Additional information

Used simplifier rules: butlastN-rec: ⊦ butlastn(n + 1, x) = butlastn(n, x.butlast);% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)