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Proof step 14 from proof datasize-append-h-new (previous step)

n < # buf1, ∀ buf1, buf0. n ≤ # buf1 → # buf1 = 0 ∨ datasize(buf1, n) = 0 ∨ datasize(buf0 + buf1, n + # buf0) = # buf0 + datasize(buf1, n) ∨ n = 0 ⊦ (buf1[n] = empty ⊃ datasize(buf0 + buf1, n + # buf0);(n + # buf0) + 1) = # buf0 + (buf1[n] = empty ⊃ datasize(buf1, n);n + 1), (buf1[n] = empty ⊃ datasize(buf1, n);n + 1) = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 15:
buf1[n] = empty, n < # buf1, ∀ buf1, buf0. n ≤ # buf1 → # buf1 = 0 ∨ datasize(buf1, n) = 0 ∨ datasize(buf0 + buf1, n + # buf0) = # buf0 + datasize(buf1, n) ∨ n = 0 ⊦ (buf1[n] = empty ⊃ datasize(buf1, n);n + 1) = 0, datasize(buf0 + buf1, n + # buf0) = # buf0 + (buf1[n] = empty ⊃ datasize(buf1, n);n + 1)

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)