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Proof step 5 from proof neq-07 (previous step)

y1 + a0 ' + y0 = y2 + a2 ' + a1 ' + y ⊦ a2 = a1, a1 = a0, a2 = a0, a1 ∈ y0, a2 ∈ y1

User applied rule structural induction

And obtained the following new goals

Proof step 6:
∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3, (a + y3) + a0 ' + y0 = y2 + a2 ' + a1 ' + y ⊦ a2 = a1, a1 = a0, a2 = a0, a1 ∈ y0, a2 ∈ a + y3

Rule argument

indvar: y1

Additional information

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