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

⊦ range=(ar0, n0, ar1, n1, 0 + m1) ↔ range=(ar0, n0, ar1, n1, 0) ∧ range=(ar0, n0 + 0, ar1, n1 + 0, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ range=(ar0, n0, ar1, n1, 0) ↔ n0 ≤ # ar0 ∧ n1 ≤ # ar1
Proof step 4:
range=(ar0, n0, ar1, n1, m1) ⊦ n0 ≤ # ar0 ∧ n1 ≤ # ar1

Rule argument

Additional information

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