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Proof step 22 from proof datasize-ls-samedata-h (previous step)

# buf0 = # buf1, m0 + 1 < (buf1[m0] = empty ⊃ datasize(buf1, m0);m0 + 1), m0 < # buf1, range=(buf0, 0, buf1, 0, datasize(buf1, # buf1)), ∀ buf, buf2. # buf ≠ # buf2 ∨ ¬ datasize(buf, m0) < datasize(buf2, m0) ∨ ¬ range=(buf, 0, buf2, 0, datasize(buf2, # buf2)) ∨ ¬ m0 ≤ # buf2 ⊦ buf0[m0] = empty

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 23:
n ≠ 0, m ≤ # buf ⊦ ¬ m + n < datasize(buf, m)

Rule argument

pos: left 2

Additional information

Used simplifier rules: datasize-ls-add: n ≠ 0, m ≤ # buf ⊦ ¬ m + n < datasize(buf, m);% ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → m ≤ n; (s)