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Proof step 2 from proof equal-add-in (previous step)

∀ a1. a1 = a ∨ a1 ∈ s ↔ a1 = b ∨ a1 ∈ s0, ∀ a1. a1 = b ∨ a1 ∈ s0 ↔ a1 = a0 ∨ a1 ≠ c ∧ a1 ∈ s1 ⊦ b = a0, b ≠ c ∧ b ∈ s1

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: b

Additional information

Used terms: b