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Proof step 12 from proof del-addeq (previous step)

a0 ∈ s, a1 ∈ s, ∀ a0. a0 ≠ b ∧ a0 ∈ s ↔ a0 ∈ s0 ⊦ a0 = b, a = b, a ∈ s, a ∈ s0, a0 ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a0

Additional information

Used terms: a0