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Proof step 2 from proof neq-01 (previous step)

y + c ' + z = y1 + a ' + x ⊦ a = c, a ∈ z, a ∈ y

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0, (a0 + y0) + c ' + z = y1 + a ' + x ⊦ a = c, a ∈ z, a ∈ a0 + y0

Rule argument

indvar: y

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 [], +;;