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Proof step 4 from proof del-addeq (previous step)

∀ a0. a0 ≠ b ∧ (a0 = a ∨ a0 ∈ s) ↔ a0 = a ∨ a0 ∈ s0 ⊦ a0 ≠ b ∧ a0 ∈ s ↔ a0 ∈ s0, a = b, a ∈ s0, a ∈ s

User applied rule case distinction

And obtained the following new goals

Proof step 5:
a0 ≠ b ∧ a0 ∈ s, ∀ a0. a0 ≠ b ∧ (a0 = a ∨ a0 ∈ s) ↔ a0 = a ∨ a0 ∈ s0 ⊦ a0 ∈ s0, a = b, a ∈ s0, a ∈ s
Proof step 7:
a0 ∈ s0, ∀ a0. a0 ≠ b ∧ (a0 = a ∨ a0 ∈ s) ↔ a0 = a ∨ a0 ∈ s0 ⊦ a0 ≠ b ∧ a0 ∈ s, a = b, a ∈ s0, a ∈ s

Rule argument

pos: right 1

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