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Proof step 3 from proof range-eq-bounds (previous step)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
⊦ range=(ar0, n0, ar1, n1, 0) ↔ n0 ≤ # ar0 ∧ n1 ≤ # ar1
Proof step 5:
range=(ar0, n0, ar1, n1, m0 +1), ∀ ar0, n0, ar1, n1. range=(ar0, n0, ar1, n1, m0) → n0 + m0 ≤ # ar0 ∧ n1 + m0 ≤ # ar1 ⊦ n0 + m0 +1 ≤ # ar0 ∧ n1 + m0 +1 ≤ # ar1

Rule argument

term: m

Additional information

Used simplifier rules: range-eq-0: ⊦ range=(ar0, n0, ar1, n1, 0) ↔ n0 ≤ # ar0 ∧ n1 ≤ # ar1;%nat freely generated by 0, +1;;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic