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Proof step 6 from proof datasize-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < # buf, m ≤ # buf, m ≤ n + 1 ⊦ buf[n] = empty, m = 0, ∃ n. m ≤ n + 1 ∧ n < # buf ∧ buf[n] ≠ empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ m; (s)% ⊦ n ≥ m ↔ m ≤ n; (s)