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Proof step 8 from proof datasize-append-empty-h (previous step)

m0 < # buf0 + # buf1, # buf0 ≤ m0 + 1 → datasize(buf1, m0 + 1 - # buf0) = 0, ∀ buf0, buf1. m0 ≤ # buf0 + # buf1 ∧ (# buf0 ≤ m0 → datasize(buf1, m0 - # buf0) = 0) → datasize(buf0 + buf1, m0) = datasize(buf0, min(m0, # buf0)) ⊦ ((buf0 + buf1)[m0] = empty ⊃ datasize(buf0 + buf1, m0);m0 + 1) = datasize(buf0, min(m0 + 1, # buf0))

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 < # buf0 + # buf1, # buf0 ≤ m0 + 1 → datasize(buf1, m0 + 1 - # buf0) = 0, ∀ buf0, buf1. m0 ≤ # buf0 + # buf1 ∧ (# buf0 ≤ m0 → datasize(buf1, m0 - # buf0) = 0) → datasize(buf0 + buf1, m0) = datasize(buf0, min(m0, # buf0)) ⊦ ((buf0 + buf1)[m0] = empty ⊃ datasize(buf0 + buf1, m0);m0 + 1) = datasize(buf0, (m0 + 1 < # buf0 ⊃ m0 + 1;# buf0))

Rule argument

spec: min-max name: min seq: ⊦ min(n, m) = (n < m ⊃ n;m) subst: [m, n] → [# buf0, m0 + 1] flags: <all paths>

Additional information

Used terms: # buf0, m0 + 1