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Proof step 2 from proof in-free-h (previous step)

m +1 ≤ # wlar, ∀ wlar, n. m ≤ # wlar → (n ∈ free-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = free ∧ wlar[m1].ec = n)) ⊦ n ∈ free-ecs(wlar, m +1) ↔ (∃ m1. m1 < m +1 ∧ wlar[m1].status = free ∧ wlar[m1].ec = n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < # wlar, ∀ wlar, n. m ≤ # wlar → (n ∈ free-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = free ∧ wlar[m1].ec = n)) ⊦ n ∈ (wlar[m].status = free ⊃ free-ecs(wlar, m) ++ wlar[m].ec;free-ecs(wlar, m)) ↔ (∃ m0. ¬ m < m0 ∧ wlar[m0].status = free ∧ wlar[m0].ec = n)

Rule argument

Additional information

Used simplifier rules: free-ecs-rec: ⊦ n < # wlar → free-ecs(wlar, n + 1) = (wlar[n].status = free ⊃ free-ecs(wlar, n) ++ wlar[n].ec ;free-ecs(wlar, n) );% ⊦ 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)