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Proof step 3 from proof delete-shift (previous step)

a ∈ s, a0 ∈ s ⊦ (∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 ∈ s0) ↔ (∀ a0. a0 ∈ s ↔ a0 ∈ s0 ++ a) ∧ ¬ a ∈ s0

User applied rule case distinction

And obtained the following new goals

Proof step 4:
a0 ∈ s, a ∈ s, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 ∈ s0 ⊦ (∀ a0. a0 ∈ s ↔ a0 ∈ s0 ++ a) ∧ ¬ a ∈ s0
Proof step 9:
a0 ∈ s, a ∈ s, ∀ a0. a0 ∈ s ↔ a0 ∈ s0 ++ a ⊦ a ∈ s0, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 ∈ s0

Rule argument

pos: right 1

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