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Proof step 7 from proof lastN-in-01 (previous step)

n0 < # x, a ∈ lastn((n0 + 1)-1, x.butlast) + x.last, ∀ a, x. a ∈ lastn(n0, x) ∧ n0 ≤ # x → a ∈ x ⊦ n0 + 1 = 0, x = [], a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 < # x, a ∈ lastn(n0, x.butlast), ∀ a, x. a ∈ lastn(n0, x) ∧ n0 ≤ # x → a ∈ x ⊦ a = x.last, x = [], a ∈ x

Rule argument

Additional information

Used simplifier rules: ⊦ x.last ∈ x ↔ x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ x + a = x + a '; (s)% ⊦ (m + 1)-1 = m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)