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Proof step 7 from proof datasize-le-h (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
m1 < # buf, m ≤ (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1), ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0 ⊦ m = 0, ∃ n. m ≤ n + 1 ∧ ¬ m1 < n ∧ buf[n] ≠ empty
Proof step 14:
m1 < # buf, ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0, ∃ n. m ≤ n + 1 ∧ ¬ m1 < n ∧ buf[n] ≠ empty ⊦ m = 0, m ≤ (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1)

Rule argument

pos: right 1

Proof step 7 from proof datasize-le-h (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information