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Proof step 3 from proof in-ecs-n (previous step)

n ≤ # wlar ⊦ m ∈ ecs(wlar, n) ↔ (∃ n0. n0 < n ∧ wlar[n0].ec = m)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n0 +1 ≤ # wlar, ∀ wlar, m. n0 ≤ # wlar → (m ∈ ecs(wlar, n0) ↔ (∃ n1. n1 < n0 ∧ wlar[n1].ec = m)) ⊦ m ∈ ecs(wlar, n0 +1) ↔ (∃ n1. n1 < n0 +1 ∧ wlar[n1].ec = m)

Rule argument

indvar: n

Additional information

Used simplifier rules: ecs-n-base: ⊦ ecs(wlar, 0) = ∅;%nat freely generated by 0, +1;;% ⊦ ¬ n ∈ ∅; (s)% ⊦ ¬ m < 0; (s)