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Proof step 4 from proof datasize-le (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

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

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] → [# buf, buf, m] flags: <all paths>

Additional information

Used simplifier rules: ⊦ n ≥ m ↔ m ≤ n; (s)% ⊦ m ≤ m; (s) Used terms: # buf, buf, m