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Proof step 30 from proof datasize-append-empty-h (previous step)

m0 + 1 = n0 + # buf0, (buf0 + buf1)[m0] = empty, datasize(buf1, n0) = 0, datasize(buf0 + buf1, m0) = datasize(buf0, (m0 < # buf0 ⊃ m0;# buf0)), m0 < # buf0 + # buf1, # buf0 ≤ m0 + 1, n0 ≤ # buf1 ⊦ datasize(buf0 + buf1, m0) = datasize(buf0, # buf0)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 31:
m ≤ # buf0 ⊦ datasize(buf0 + buf1, m) = datasize(buf0, m)
Proof step 32:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)

Rule argument

pos: left 4

Additional information

Used simplifier rules: datasize-append-lower-h: m ≤ # buf0 ⊦ datasize(buf0 + buf1, m) = datasize(buf0, m);%datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);% ⊦ m < n → m ≤ n; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ n < # buf → (buf + buf0)[n] = buf[n]; (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ m + 1 ≠ 0; (s)