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Proof step 15 from proof isempty-ext (previous step)

n0 + n < # buf, isempty(buf, n0, n), ∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty ⊦ buf[n0 + n] = empty ↔ (∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty)

User applied rule case distinction

And obtained the following new goals

Proof step 16:
buf[n0 + n] = empty, n0 + n < # buf, isempty(buf, n0, n), ∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty ⊦ ∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty
Proof step 18:
n0 + n < # buf, isempty(buf, n0, n), ∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty, ∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty ⊦ buf[n0 + n] = empty

Rule argument

pos: right 1

Proof step 15 from proof isempty-ext (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information