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Proof step 7 from proof neq-07 (previous step)

(a ' + y3) + a0 ' + y0 = y2 + a2 ' + a1 ' + y, a ∈ y2, ∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3 ⊦ a2 = a, a2 = a0, a1 = a0, a2 = a1, a2 ∈ y3, a1 ∈ y0

User applied rule structural induction

And obtained the following new goals

Proof step 8:
∀ a, y3, a0, y0, a2, a1, y. (a ' + y3) + a0 ' + y0 = y1 + a2 ' + a1 ' + y ∧ a ∈ y1 ∧ (∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3) → a2 = a ∨ a2 = a0 ∨ a1 = a0 ∨ a2 = a1 ∨ a2 ∈ y3 ∨ a1 ∈ y0, (a ' + y3) + a0 ' + y0 = (a3 + y1) + a2 ' + a1 ' + y, a ∈ a3 + y1, ∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3 ⊦ a2 = a, a2 = a0, a1 = a0, a2 = a1, a2 ∈ y3, a1 ∈ y0

Rule argument

indvar: y2

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 [], +;;