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Proof step 2 from proof range-eq-cut (previous step)

n1 + m ≤ # ar1 ⊦ range=(ar0, n0, ar1, n1, m) ↔ range=(ar0, n0, resize(ar1, n1 + m), n1, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ n ≤ # ar1 → (range=(ar0, n0, resize(ar1, n), n1, m) ↔ range=(ar0, n0, ar1, n1, m) ∧ m + n1 ≤ n)

Rule argument

Additional information

Used simplifier rules: range-eq-resize: ⊦ n ≤ # ar1 → ( range=(ar0, n0, resize(ar1, n), n1, m) ↔ range=(ar0, n0, ar1, n1, m) ∧ m + n1 ≤ n);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)