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Proof step 4 from proof in-ecs-n (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < # wlar, ∀ wlar, m. n0 ≤ # wlar → (m ∈ ecs(wlar, n0) ↔ (∃ n1. n1 < n0 ∧ wlar[n1].ec = m)) ⊦ m ∈ ecs(wlar, n0) ++ wlar[n0].ec ↔ (∃ n. ¬ n0 < n ∧ wlar[n].ec = m)

Rule argument

Additional information

Used simplifier rules: ecs-n-rec: ⊦ n < # wlar → ecs(wlar, n + 1) = ecs(wlar, n) ++ wlar[n].ec;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)