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Proof step 2 from proof del-feq-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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)