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Proof step 2 from proof k-delete-rec (previous step)

⊦ (s ++ b) -- a = (s -- a) ++ b, a = b, b ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a0 = b ∨ a0 ≠ a ∧ a0 ∈ s ⊦ a = b, b ∈ s, a0 ≠ a ∧ (a0 = b ∨ a0 ∈ s)

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)