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Proof step 12 from proof firstoc (previous step)

∀ a1. a1 ∈ y ∨ a1 ∈ a ' ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0), a = a0 ∨ a ∈ a ' ∨ a ∈ z ⊦ a ∈ y, ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0, a0 ' + y + a ' + z = (a0 ' + y) + a ' + z ∧ ¬ a ∈ a0 ' + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
∀ a1. a1 ∈ y ∨ a1 ∈ a0 ' ∨ a1 ∈ z → (∃ y0, z0. y + a0 ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0) ⊦ a0 ∈ y, ∃ y0, z0. a0 ' + y + a0 ' + z = y0 + a0 ' + z0 ∧ ¬ a0 ∈ y0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%in-rec: ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x;