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Proof step 8 from proof size-one-02 (previous step)

# s0 = 1, a0 ∈ s0, a ∈ s0, b ∈ s0, ∀ b0, a1. b0 ∈ s0 ∧ a1 ∈ s0 → a1 = b0 ⊦ a = b

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: b, a

Additional information

Used terms: b, a