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Proof step 1 from proof eq-01 (conclusion)

⊦ x + a ' + y = b ' + z ∧ x + a ' + y0 = b0 ' + z0 → b = b0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + a ' + y0 = b0 ' + z0, x + a ' + y = b ' + z, b0 ∈ x ∨ b0 = a ∨ b0 ∈ y0, b ∈ x ∨ b = a ∨ b ∈ y ⊦ b = b0

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-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);