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Proof step 2 from proof del (previous step)

a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
a ∈ s0 ++ a0, ∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1) ⊦ ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

Rule argument

indvar: s

Additional information

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