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Proof step 36 from proof size-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a0 = a ∨ a0 ∈ s0, b = a ∨ b ∈ s0 ⊦ a1 ∈ s0 ↔ a1 = b, 1 = # s0, a0 = b, s0 = ∅ ++ a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

Additional information

Used simplifier rules: in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;