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Proof step 3 from proof append-size-less (previous step)

x + x0 = x1 + x2, # x < # x1 ⊦ x1 = [], ∃ y. x1 = x + y ∧ x0 = y + x2

User applied rule structural induction

And obtained the following new goals

Proof step 4:
(a + x3) + x0 = x1 + x2, #(a + x3) < # x1, ∀ x0, x1, x2. x3 + x0 = x1 + x2 ∧ # x3 < # x1 → x1 = [] ∨ (∃ y. x1 = x3 + y ∧ x0 = y + x2) ⊦ x1 = [], ∃ y. x1 = (a + x3) + y ∧ x0 = y + x2

Rule argument

indvar: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ # x = 0 ↔ x = []; (s)