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Proof step 8 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, min(# buf, m0 + 1))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ min(n, m) = (n < m ⊃ n;m)
Proof step 10:
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))

Rule argument

spec: min-max name: min seq: ⊦ min(n, m) = (n < m ⊃ n;m) subst: [m, n] → [m0 + 1, # buf] flags: <all paths>

Additional information

Used terms: m0 + 1, # buf