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Proof step 2 from proof neq-07 (previous step)

y1 + a0 ' + y0 + a ' = y2 + a ' + a1 ' + z0 ⊦ a = a1, a = a0, a1 = a0, a1 ∈ y0, a ∈ y1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
y1 + a0 ' + y0 + a ' = y2 + a ' + a1 ' + z0 ⊦ a = a1, a = a0, a1 = a0, a1 ∈ y0, a ∈ y1

Rule argument

Additional information

Used simplifier rules: in-16: ⊦ x + a ' = y + b ' + z ∧ a ≠ b → a ∈ z;%a: ⊦ (x + y) + z = x + y + z;%in-15: ⊦ x + a ' = y → a ∈ y;%in-16: ⊦ x + a ' = y + b ' + z ∧ a ≠ b → a ∈ z;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;%in-17: ⊦ a ∈ b ' ↔ a = b;