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Proof step 1 from proof Del-in (conclusion)

⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ a, b. a ∈ st0 -- b ↔ a ≠ b ∧ a ∈ st0 ⊦ a ∈ st0[a0, d] -- b ↔ a ≠ b ∧ a ∈ st0[a0, d]

Rule argument

indvar: st

Additional information

Used simplifier rules: gen: ⊦ ;%Del-empty: ⊦ ∅ -- a = ∅;%In-empty: ⊦ ¬ a ∈ ∅;