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Proof step 8 from proof rangeeq-subarray (previous step)

n0 < m0, 2 * n + 5 + m0 ≤ # avol0, 2 * n + 5 + m0 ≤ # avol, ∀ n2. n2 < n + m0 → avol[(5 + n) + n2] = avol0[(5 + n) + n2] ⊦ avol[(5 + 2 * n) + n0] = avol0[(5 + 2 * n) + n0]

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: n + n0 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + m = 2 * m; (s) Used terms: n + n0