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Proof step 43 from proof size-two (previous step)

a2 = a ∨ a2 ∈ s0 ↔ a2 = b ∨ a2 = a0, a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a1 ∈ s0, ∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, b = a ∨ b ∈ s0, a0 = a ∨ a0 ∈ s0 ⊦ a2 ∈ s0 ↔ a2 = a0, 1 = # s0, a0 = b, a1 = b, a ∈ s0, ∃ a. s0 = ∅ ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
b ∈ s0, a0 ∈ s0, ∀ a1. a1 = a ∨ a1 ∈ s0 → a1 = b ∨ a1 = a0 ⊦ 1 = # s0, b = a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);