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Proof step 41 from proof datasize-append-empty-h (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 42:
m0 + 1 = n0 + # buf0, datasize(buf1, n0) = 0, m0 < # buf0 + # buf1, # buf0 ≤ m0 + 1, ∀ buf0, buf1. m0 ≤ # buf0 + # buf1 ∧ (# buf0 ≤ m0 → datasize(buf1, m0 - # buf0) = 0) → datasize(buf0 + buf1, m0) = datasize(buf0, min(m0, # buf0)) ⊦ m0 + 1 = datasize(buf0, # buf0), (buf0 + buf1)[m0] = empty

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0 + 1, # buf0, n]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0 + 1, # buf0, n