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Proof step 9 from proof neq-07 (previous step)

y3 + a0 ' + y0 = y1 + a2 ' + a1 ' + y, ∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3, ∀ a, y3, a0, y0, a2, a1, y. (a ' + y3) + a0 ' + y0 = y1 + a2 ' + a1 ' + y ∧ a ∈ y1 ∧ (∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3) → a2 = a ∨ a2 = a0 ∨ a1 = a0 ∨ a2 = a1 ∨ a2 ∈ y3 ∨ a1 ∈ y0 ⊦ a2 = a1, a1 = a0, a2 = a0, a2 = a, a1 ∈ y0, a2 ∈ y3

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a0, y0, y1, a2, a1, y abort: F discard: T computed: T

Additional information

Used terms: a0, y0, y1, a2, a1, y