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Proof step 29 from proof size-two (previous step)

∀ a1. a1 ∈ s0 ++ a ↔ a1 ∈ ∅ ++ a0 ++ b ⊦ 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 30:
∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0 ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, s0 = ∅ ++ a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;