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Proof step 1 from proof inex-01 (conclusion)

⊦ a ' + x = y + b ' + z ∧ a ≠ b → (∃ y0. y = a ' + y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ' + x = y + b ' + z, a ∈ y ⊦ a = b, ∃ y0. y = a ' + y0

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);%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x);%inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0);