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Proof step 20 from proof uniquelen (previous step)

a0 ∈ ax0, unique_al(ax0), unique_al(ay0 + (a × d)' + az), ∀ ay0. (∀ a. a ∈ ax0 ↔ a ∈ ay0) ∧ unique_al(ay0) → # ax0 = # ay0, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦ # ax0 = # ay0 + # az, a ∈ ax0, a ∈ ay0, a ∈ az, a0 ∈ ay0, a0 ∈ az

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: a0

Additional information

Used terms: a0