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Proof step 2 from proof lastn-neq-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
lastn(n0 + 1, x) = b ' + y, a ∈ y, n0 < # x, ∀ x, a, y, b. n0 ≤ # x ∧ a ∈ y ∧ ¬ a ∈ x → lastn(n0, x) ≠ b ' + y ⊦ x = [], a ∈ x

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2