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Proof step 2 from proof inapp-03 (previous step)

x + a ' + z = y + b ' + y0 ⊦ b ∈ x, a ∈ y, a = b ∧ x = y ∧ z = y0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a, z, y, b, y0. x0 + a ' + z = y + b ' + y0 → b ∈ x0 ∨ a ∈ y ∨ a = b ∧ x0 = y ∧ z = y0, (a0 + x0) + a ' + z = y + b ' + y0 ⊦ b ∈ a0 + x0, a ∈ y, a = b ∧ a0 + x0 = y ∧ z = y0

Rule argument

indvar: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0);%in-empty: ⊦ ¬ a ∈ [];%list freely generated by [], +;;