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Proof step 6 from proof datasize-ls-samedata-h (previous step)

# buf0 = # buf1, datasize(buf0, m0 +1) < datasize(buf1, m0 +1), range=(buf0, 0, buf1, 0, datasize(buf1, # buf1)), m0 +1 ≤ # buf1, ∀ buf0, buf1. ¬(# buf0 = # buf1 ∧ datasize(buf0, m0) < datasize(buf1, m0) ∧ range=(buf0, 0, buf1, 0, datasize(buf1, # buf1)) ∧ m0 ≤ # buf1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 8:
# 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 ⊦

Rule argument

Additional information

Used simplifier rules: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)