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Proof step 6 from proof two (previous step)

a ' + b ' = x + a0 ' + c ' + a1 ' + y0, b ∈ x, a ∈ x, ∀ a2, b0, x0, a3, c0. a2 ' + b0 ' = x0 + a3 ' + c0 ' + y0 ∧ (a2 ∈ x0 ∨ a2 = a3 ∨ a2 = c0 ∨ a2 ∈ y0) → x0 = [] ∧ y0 = [] ∧ a2 = a3 ∧ b0 = c0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a, b, x + a0 ', c, a1

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;%e-02: ⊦ x + a ' ≠ []; Used terms: a, b, x + a0 ', c, a1