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Proof step 1 from proof put-in (conclusion)

⊦ a ≠ x[m] ∧ m < # x → (a ∈ x[m, b] ↔ a = b ∨ a ∈ x)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ a, m, b. a ≠ x0[m] ∧ m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ⊦ a ≠ (a0 + x0)[m] ∧ m < #(a0 + x0) → (a ∈ (a0 + x0)[m, b] ↔ a = b ∨ a ∈ a0 + x0)

Rule argument

indvar: x

Additional information

Used simplifier rules: put-in-02: ⊦ m < # x → a ∈ x[m, a];%list freely generated by [], +;;%in-empty: ⊦ ¬ a ∈ []; from specification list%size-01: ⊦ # [] = 0; from specification list-data%zero-01: ⊦ ¬ m < 0; from specification nat-basic1