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Proof step 8 from proof lasttwo (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ a0 = a, a ∈ z0, a0 ∈ y0, a0 ∈ z0, ∃ b, y1, z1. y + a0 ' + y0 + a ' + z0 = y1 + b ' + z1 ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z1 ∧ ¬ a0 ∈ z1

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%in-13: ⊦ x = y + a ' + z → a ∈ x;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);