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Proof step 2 from proof inapp (previous step)

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

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%inapp-04: ⊦ ¬ a ∈ x → (a ' + x = y + a ' + z ↔ y = [] ∧ z = x);%in-empty: ⊦ ¬ a ∈ [];%list freely generated by [], +;;