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Proof step 4 from proof neq-01 (previous step)

(a0 ' + y0) + c ' + z = y1 + a ' + x, a0 ∈ y1, ∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0 ⊦ a = a0, a = c, a ∈ y0, a ∈ z

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: y1

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 [], +;;