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Proof step 4 from proof rangeeq-write-right-concrete-below (previous step)

n0 < n ⊦ (n + m) + m1 ≤ # avol ∧ (n + m) + m1 ≤ # avol0[n0, aleb] ∧ (∀ n2. n2 < m1 → avol[(n + m) + n2] = avol0[n0, aleb][(n + m) + n2]) ↔ range=(avol, n + m, avol0, n + m, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < n ⊦ (n + m) + m1 ≤ # avol ∧ (n + m) + m1 ≤ # avol0 ∧ (∀ n1. n1 < m1 → avol[(n + m) + n1] = avol0[(n + m) + n1]) ↔ range=(avol, n + m, avol0, n + m, m1)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%m ≠ n ⊦ avol[m, aleb][n] = avol[n]; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ # avol[n, aleb] = # avol; (s)