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Proof step 6 from proof in-used-h (previous step)

n ∈ used-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = used ∧ wlar[m1].ec = n), wlar[m] = wl-entry(m2, stat), m < # wlar ⊦ n ∈ (stat = used ⊃ used-ecs(wlar, m) ++ m2;used-ecs(wlar, m)) ↔ (∃ m0. ¬ m < m0 ∧ wlar[m0].status = used ∧ wlar[m0].ec = n)

User applied rule case distinction

And obtained the following new goals

Proof step 7:
n ∈ used-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = used ∧ wlar[m1].ec = n), wlar[m] = wl-entry(m2, stat), m < # wlar, n ∈ (stat = used ⊃ used-ecs(wlar, m) ++ m2;used-ecs(wlar, m)) ⊦ ∃ m0. ¬ m < m0 ∧ wlar[m0].status = used ∧ wlar[m0].ec = n
Proof step 21:
n ∈ used-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = used ∧ wlar[m1].ec = n), wlar[m] = wl-entry(m2, stat), m < # wlar, ∃ m0. ¬ m < m0 ∧ wlar[m0].status = used ∧ wlar[m0].ec = n ⊦ n ∈ (stat = used ⊃ used-ecs(wlar, m) ++ m2;used-ecs(wlar, m))

Rule argument

pos: right 1

Proof step 6 from proof in-used-h (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information