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Proof step 3 from proof lastn-neq-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
lastn(n0 + 1, x) = b ' + y, n0 < # x, a ∈ y, ∀ x, b, y, a. lastn(n0, x) = b ' + y ∧ a ∈ y ∧ n0 ≤ # x → a ∈ x ⊦ x = [], a ∈ x

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%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)