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Proof step 15 from proof datasize-subarray-h (previous step)

m0 + m1 = # buf, buf[m0 + m2] = empty, datasize(buf, m0 + m2) = n0 + m0, m2 < m1, ∀ m0, m1, buf. m0 + m1 = # buf ∧ m2 ≤ m1 → datasize(subarray(buf, m0, m1), m2) = (datasize(buf, m0 + m2) ≤ m0 ⊃ 0;datasize(buf, m0 + m2) - m0) ⊦ datasize(subarray(buf, m0, m1), m2) = n0, n0 = 0

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: m0, m1, buf abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n - m = n; (s) Used terms: m0, m1, buf