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Proof step 3 from proof lasttwo (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ y ∨ a = a0 ∨ a ∈ z ⊦ a0 ∈ z, ∃ b, y0, z0. y + a0 ' + z = y0 + b ' + z0 ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z0 ∧ ¬ a0 ∈ z0

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);