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

s ++ a = s0 ++ b, s0 ++ b = (s1 -- c) ++ a0 ⊦ b = a0, b ∈ s1 -- c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a1. a1 = b ∨ a1 ∈ s0 ↔ a1 = a0 ∨ a1 ≠ c ∧ a1 ∈ s1, ∀ a1. a1 = a ∨ a1 ∈ s ↔ a1 = b ∨ a1 ∈ s0 ⊦ b = a0, b ≠ c ∧ b ∈ s1

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)