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Proof step 44 from proof size-two (previous step)

b ∈ s0, a0 ∈ s0, ∀ a1. a1 = a ∨ a1 ∈ s0 → a1 = b ∨ a1 = a0 ⊦ 1 = # s0, b = a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used terms: a