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Proof step 2 from proof datasize-ls-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m0 ≤ # buf ⊦ datasize(buf, m0) < m ↔ ¬(∃ n. m ≤ n + 1 ∧ n < m0 ∧ buf[n] ≠ empty), m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ n ≥ m ↔ m ≤ n; (s)