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Proof step 16 from proof datasize-append-h-new (previous step)

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 ⊦ datasize(buf0 + buf1, n + # buf0) = # buf0 + datasize(buf1, n), datasize(buf1, n) = 0

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: buf1, buf0

Additional information

Used simplifier rules: datasize-append-lower-h: m ≤ # buf0 ⊦ datasize(buf0 + buf1, m) = datasize(buf0, m);%datasize-base: ⊦ datasize(buf, 0) = 0;% ⊦ m < n → m ≤ n; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic Used terms: buf1, buf0