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Proof step 8 from proof datsize-write-higher-h (previous step)

((m1 < m ∨ m + n ≤ m1 ⊃ buf[m1];buf0[m1 - m]) = empty ⊃ datasize(copy(buf0, 0, buf, m, n), m1);m1 + 1) < (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1), m1 < # buf, n ≤ # buf0, m + n ≤ # buf, datasize(buf, # buf) ≤ m, ∀ buf1, buf2, m0, n0. ¬ datasize(copy(buf1, 0, buf2, m0, n0), m1) < datasize(buf2, m1) ∨ ¬ datasize(buf2, # buf2) ≤ m0 ∨ ¬ m0 + n0 ≤ # buf2 ∨ ¬ n0 ≤ # buf1 ∨ ¬ m1 ≤ # buf2 ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
(m1 < m ∨ m + n ≤ m1 ⊃ buf[m1];buf0[m1 - m]) = empty, m1 < # buf, datasize(copy(buf0, 0, buf, m, n), m1) < (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1), datasize(buf, # buf) ≤ m, m + n ≤ # buf, n ≤ # buf0, ∀ buf1, buf2, m0, n0. ¬ datasize(copy(buf1, 0, buf2, m0, n0), m1) < datasize(buf2, m1) ∨ ¬ datasize(buf2, # buf2) ≤ m0 ∨ ¬ m0 + n0 ≤ # buf2 ∨ ¬ n0 ≤ # buf1 ∨ ¬ m1 ≤ # buf2 ⊦
Proof step 23:
m1 < # buf, m1 + 1 < (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1), datasize(buf, # buf) ≤ m, m + n ≤ # buf, n ≤ # buf0, ∀ buf1, buf2, m0, n0. ¬ datasize(copy(buf1, 0, buf2, m0, n0), m1) < datasize(buf2, m1) ∨ ¬ datasize(buf2, # buf2) ≤ m0 ∨ ¬ m0 + n0 ≤ # buf2 ∨ ¬ n0 ≤ # buf1 ∨ ¬ m1 ≤ # buf2 ⊦ (m1 < m ∨ m + n ≤ m1 ⊃ buf[m1];buf0[m1 - m]) = empty

Rule argument

pos: left 1

Proof step 8 from proof datsize-write-higher-h (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information