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Proof step 6 from proof datasize-append-empty-h (previous step)

m0 +1 ≤ # buf0 + # buf1, # buf0 ≤ m0 +1 → datasize(buf1, m0 +1 - # buf0) = 0, ∀ buf0, buf1. m0 ≤ # buf0 + # buf1 ∧ (# buf0 ≤ m0 → datasize(buf1, m0 - # buf0) = 0) → datasize(buf0 + buf1, m0) = datasize(buf0, min(m0, # buf0)) ⊦ datasize(buf0 + buf1, m0 +1) = datasize(buf0, min(m0 +1, # buf0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 8:
m0 < # buf0 + # buf1, # buf0 ≤ m0 + 1 → datasize(buf1, m0 + 1 - # buf0) = 0, ∀ buf0, buf1. m0 ≤ # buf0 + # buf1 ∧ (# buf0 ≤ m0 → datasize(buf1, m0 - # buf0) = 0) → datasize(buf0 + buf1, m0) = datasize(buf0, min(m0, # buf0)) ⊦ ((buf0 + buf1)[m0] = empty ⊃ datasize(buf0 + buf1, m0);m0 + 1) = datasize(buf0, min(m0 + 1, # buf0))

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% ⊦ 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)