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Proof step 3 from proof datasize-ls-h (previous step)

m0 ≤ # buf ⊦ datasize(buf, m0) < m ↔ ¬(∃ n. m ≤ n + 1 ∧ n < m0 ∧ buf[n] ≠ empty), m = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ m0 ≤ # buf ∧ m ≠ 0 → (m ≤ datasize(buf, m0) ↔ (∃ n. n + 1 ≥ m ∧ n < m0 ∧ buf[n] ≠ empty))
Proof step 5:
m0 ≤ # buf, m0 ≤ # buf ∧ m ≠ 0 → (m ≤ datasize(buf, m0) ↔ (∃ n. n + 1 ≥ m ∧ n < m0 ∧ buf[n] ≠ empty)) ⊦ datasize(buf, m0) < m ↔ ¬(∃ n. m ≤ n + 1 ∧ n < m0 ∧ buf[n] ≠ empty), m = 0

Rule argument

spec: <toplevel> name: datasize-le-h seq: ⊦ m0 ≤ # buf ∧ m ≠ 0 → ( m ≤ datasize(buf, m0) ↔ (∃ n. n + 1 ≥ m ∧ n < m0 ∧ buf[n] ≠ empty)) subst: [m0, buf, m] → [m0, buf, m]

Additional information

Used terms: m0, buf, m