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Proof step 2 from proof isin (previous step)

a ' + x = y + b ' + z ⊦ y = [], b ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ' + x = y + b ' + z, a ∈ y ∨ a ∈ b ' ∨ a ∈ z ⊦ y = [], b ∈ x

Rule argument

Additional information

Used simplifier rules: in-14: ⊦ a ' + x = y → a ∈ y;%a: ⊦ (x + y) + z = x + y + z;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;