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Proof step 2 from proof unequal (previous step)

# x = # y ⊦ x = y, ∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y = z + b ' + z1

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1), # x = #(a + y0) ⊦ x = a + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ x = z + a0 ' + z0 ∧ a + y0 = z + b ' + z1

Rule argument

indvar: y

Additional information

Used simplifier rules: list freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data%l: ⊦ # x = 0 ↔ x = []; from specification list-data