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Proof step 31 from proof size-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a0 = a ∨ a0 ∈ s0 ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, s0 = ∅ ++ a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

Proof step 31 from proof size-two (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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