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Proof step 4 from proof datasize-append-lower-h (previous step)

m0 +1 ≤ # buf0, ∀ buf0, buf1. m0 ≤ # buf0 → datasize(buf0 + buf1, m0) = datasize(buf0, m0) ⊦ datasize(buf0 + buf1, m0 +1) = datasize(buf0, m0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 6:
m0 < # buf0, ∀ buf0, buf1. m0 ≤ # buf0 → datasize(buf0 + buf1, m0) = datasize(buf0, m0) ⊦ (buf0[m0] = empty ⊃ datasize(buf0 + buf1, m0);m0 + 1) = (buf0[m0] = empty ⊃ datasize(buf0, m0);m0 + 1)

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%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ n < # buf → (buf + buf0)[n] = buf[n]; (s)% ⊦ m + 1 ≤ n ↔ m < n; (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)