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Proof step 20 from proof datasize-ls-samedata-h (previous step)

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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: m0

Additional information

Used terms: m0