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Proof step 24 from proof datasize-further (previous step)

n < # buf, n < n1, ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)) ⊦ (buf[n] = empty ⊃ datasize(buf, n);n + 1) = datasize(buf, n1) ↔ (∀ n0. n < n0 ∧ n0 < n1 → buf[n0] = empty)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 25:
(buf[n] = empty ⊃ datasize(buf, n);n + 1) = datasize(buf, n1), n < n1, n < # buf, ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)) ⊦ ∀ n0. n < n0 ∧ n0 < n1 → buf[n0] = empty
Proof step 48:
n < n1, n < # buf, ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)), ∀ n0. n < n0 ∧ n0 < n1 → buf[n0] = empty ⊦ (buf[n] = empty ⊃ datasize(buf, n);n + 1) = datasize(buf, n1)

Rule argument

pos: right 1

Proof step 24 from proof datasize-further (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information