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Proof step 7 from proof rangeeq-emptyleb-datasize (previous step)

# buf = n0 + n1, n2 < n0 + n1, n0 < (buf[n2] = empty ⊃ datasize(buf, n2);n2 + 1), range=(buf, n0, emptyleb, 0, n1), ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3 ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
# buf = n0 + n1, buf[n2] = empty, n2 < n0 + n1, n0 < datasize(buf, n2), range=(buf, n0, emptyleb, 0, n1), ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3 ⊦
Proof step 9:
# buf = n0 + n1, n2 < n0 + n1, n0 < n2 + 1, range=(buf, n0, emptyleb, 0, n1), ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3 ⊦ buf[n2] = empty

Rule argument

pos: left 3

Proof step 7 from proof rangeeq-emptyleb-datasize (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

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