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Proof step 11 from proof firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ 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-10: ⊦ ¬ a ∈ x → (a ∈ b ' + x ↔ a = b);