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Proof step 2 from proof inex (previous step)

x + a ' = y + b ' + z ⊦ a = b, ∃ z0. z = z0 + a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x + a ' = y + b ' + z ⊦ a = b, ∃ z0. z = z0 + a '

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;