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Proof step 6 from proof neq-01 (previous step)

y0 + c ' + z = y + a ' + x, ∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0, ∀ a0, y0, c, z, a, x. (a0 ' + y0) + c ' + z = y + a ' + x ∧ a0 ∈ y ∧ (∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0) → a = a0 ∨ a = c ∨ a ∈ y0 ∨ a ∈ z ⊦ a = c, a = a0, a ∈ z, a ∈ y0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: c, z, y, a, x

Additional information

Used terms: c, z, y, a, x