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Proof step 5 from proof lasttwo (previous step)

a ∈ y ∨ a = a0 ∨ a ∈ z ⊦ a0 ∈ z, y + a0 ' + z = y + a0 ' + z ∧ (a0 = a ∨ a0 = a0) ∧ ¬ 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 6:
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;