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Proof step 3 from proof inapp-06 (previous step)

x + a ' + y = x0 + b ' + z ⊦ b ∈ y, a ∈ z, x = x0 ∧ a = b ∧ y = z

User applied rule structural induction

And obtained the following new goals

Proof step 4:
(a0 + x1) + a ' + y = x0 + b ' + z, ∀ a, y, x0, b, z. x1 + a ' + y = x0 + b ' + z → b ∈ y ∨ a ∈ z ∨ x1 = x0 ∧ a = b ∧ y = z ⊦ a ∈ z, b ∈ y, a0 + x1 = x0 ∧ a = b ∧ y = z

Rule argument

indvar: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%in-14: ⊦ a ' + x = y → a ∈ y;%isin: ⊦ a ' + x = y + b ' + z ∧ y ≠ [] → b ∈ x;%neq-02: ⊦ ¬ a ∈ z ∧ a ≠ c → c ' + z ≠ y1 + a ' + x;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%list freely generated by [], +;;