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Proof step 3 from proof in-01-02 (previous step)

⊦ a ∈ ins≤(b, x) ↔ a ∈ x, a = b

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ a, b. (a ∈ ins≤(b, x0) ↔ a ∈ x0) ∨ a = b ⊦ a ∈ ins≤(b, a0 + x0) ↔ a ∈ a0 + x0, a = b

Rule argument

indvar: x

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';%list freely generated by [], +;;% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ ¬ a ∈ []; (s)