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Proof step 5 from proof datasize-append-h-new (previous step)

m ≤ # buf0 + # buf1 ⊦ datasize(buf0 + buf1, m) = (m ≤ # buf0 ∨ datasize(buf1, m - # buf0) = 0 ⊃ datasize(buf0, min(m, # buf0));# buf0 + datasize(buf1, m - # buf0)), m ≤ # buf0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# buf0 < m, m ≤ # buf0 + # buf1 ⊦ # buf1 = 0, datasize(buf0 + buf1, m) = (datasize(buf1, m - # buf0) = 0 ⊃ datasize(buf0, # buf0);# buf0 + datasize(buf1, m - # buf0))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n1 < n0 → min(n0, n1) = n1; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)