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Proof step 10 from proof datasize-empbuf-h (previous step)

m0 < # buf + n, ∀ buf, n. m0 ≤ # buf + n → datasize(buf + mkempbuf(n), m0) = datasize(buf, min(# buf, m0)) ⊦ ((buf + mkempbuf(n))[m0] = empty ⊃ datasize(buf + mkempbuf(n), m0);m0 + 1) = datasize(buf, (# buf < m0 + 1 ⊃ # buf;m0 + 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m0 < # buf + n, ∀ buf, n. m0 ≤ # buf + n → datasize(buf + mkempbuf(n), m0) = datasize(buf, min(# buf, m0)) ⊦ ((buf + mkempbuf(n))[m0] = empty ⊃ datasize(buf + mkempbuf(n), m0);m0 + 1) = datasize(buf, (¬ m0 < # buf ⊃ # buf;m0 + 1))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)