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Proof step 1 from proof in-free-h (conclusion)

m0 ≤ # wlar ⊦ n ∈ free-ecs(wlar, m0) ↔ (∃ m. m < m0 ∧ wlar[m].status = free ∧ wlar[m].ec = n)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
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)

Rule argument

term: m0

Additional information

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