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Proof step 4 from proof inapp-03 (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ a0, x0, a, z, b, y0. (a0 ' + x0) + a ' + z = y1 + b ' + y0 ∧ a0 ∈ y1 ∧ (∀ a, z, y, b, y0. x0 + a ' + z = y + b ' + y0 → b ∈ x0 ∨ a ∈ y ∨ a = b ∧ x0 = y ∧ z = y0) → b = a0 ∨ b ∈ x0 ∨ a ∈ y1 ∨ a = b ∧ a0 ' + x0 = y1 ∧ z = y0, (a0 ' + x0) + a ' + z = (a1 + y1) + b ' + y0, a0 ∈ a1 + y1, ∀ a, z, y, b, y0. x0 + a ' + z = y + b ' + y0 → b ∈ x0 ∨ a ∈ y ∨ a = b ∧ x0 = y ∧ z = y0 ⊦ b = a0, b ∈ x0, a ∈ a1 + y1, a = b ∧ a0 ' + x0 = a1 + y1 ∧ z = y0

Rule argument

indvar: y

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%list freely generated by [], +;;