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Proof step 19 from proof datasize-append-empty-h (previous step)

(buf0 + buf1)[m0] = empty, 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)) ⊦ datasize(buf0 + buf1, m0) = datasize(buf0, # buf0), m0 + 1 < # buf0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
(buf0 + buf1)[m0] = empty, 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)) ⊦ datasize(buf0 + buf1, m0) = datasize(buf0, # buf0)

Rule argument

Additional information

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