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Proof step 42 from proof datasize-append-empty-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
m0 + 1 = n0 + # buf0, datasize(buf1, n0) = 0, m0 < # buf0 + # buf1, # buf0 ≤ m0 + 1, n0 ≤ # buf1, ∀ 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

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat