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Proof step 6 from proof range-eq-append (previous step)

∀ ar0, n0, ar1, n1, m1. range=(ar0, n0, ar1, n1, m + m1) ↔ range=(ar0, n0, ar1, n1, m) ∧ range=(ar0, n0 + m, ar1, n1 + m, m1) ⊦ range=(ar0, n0, ar1, n1, m +1 + m1) ↔ range=(ar0, n0, ar1, n1, m +1) ∧ range=(ar0, n0 + m +1, ar1, n1 + m +1, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ ar0, n0, ar1, n1, m1. range=(ar0, n0, ar1, n1, m + m1) ↔ range=(ar0, n0, ar1, n1, m) ∧ range=(ar0, n0 + m, ar1, n1 + m, m1) ⊦ range=(ar0, n0, ar1, n1, (m + 1) + m1) ↔ range=(ar0, n0, ar1, n1, m + 1) ∧ range=(ar0, n0 + m + 1, ar1, n1 + m + 1, m1)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)