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Proof step 8 from proof datasize-ls-samedata-h (previous step)

# buf0 = # buf1, (buf0[m0] = empty ⊃ datasize(buf0, m0);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 ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
# buf0 = # buf1, buf0[m0] = empty, m0 < # buf1, datasize(buf0, m0) < (buf1[m0] = empty ⊃ datasize(buf1, m0);m0 + 1), 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 ⊦
Proof step 21:
# buf0 = # buf1, m0 < # buf1, m0 + 1 < (buf1[m0] = empty ⊃ datasize(buf1, m0);m0 + 1), 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

Rule argument

pos: left 2

Proof step 8 from proof datasize-ls-samedata-h (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

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