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Proof step 2 from proof put-in (previous step)

m < # x ⊦ a ∈ x[m, b] ↔ a = b ∨ a ∈ x, a = x[m]

User applied rule structural induction

And obtained the following new goals

Proof step 3:
m < #(a0 + x0), ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ∨ a = x0[m] ⊦ a ∈ (a0 + x0)[m, b] ↔ a = b ∨ a ∈ a0 + x0, a = (a0 + x0)[m]

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;% ⊦ ¬ a ∈ []; (s)% ⊦ # [] = 0; (s)% ⊦ ¬ n < 0; (s)