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Proof step 5 from proof datasize-le-h (previous step)

m1 +1 ≤ # buf, ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0 ⊦ m ≤ datasize(buf, m1 +1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 +1 ∧ buf[n] ≠ empty), m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 7:
m1 < # buf, ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0 ⊦ m ≤ (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) ↔ (∃ n. m ≤ n + 1 ∧ ¬ m1 < n ∧ buf[n] ≠ empty), m = 0

Rule argument

Additional information

Used simplifier rules: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)