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Proof step 4 from proof lastN-in-01 (previous step)

a ∈ lastn(n0 +1, x), n0 +1 ≤ # x, ∀ a, x. a ∈ lastn(n0, x) ∧ n0 ≤ # x → a ∈ x ⊦ a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < # x, a ∈ lastn(n0 + 1, x), ∀ a, x. a ∈ lastn(n0, x) ∧ n0 ≤ # x → a ∈ x ⊦ x = [], a ∈ x

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)