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Proof step 1 from proof range-eq-append (conclusion)

⊦ range=(ar0, n0, ar1, n1, m0 + m1) ↔ range=(ar0, n0, ar1, n1, m0) ∧ range=(ar0, n0 + m0, ar1, n1 + m0, m1)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
⊦ range=(ar0, n0, ar1, n1, 0 + m1) ↔ range=(ar0, n0, ar1, n1, 0) ∧ range=(ar0, n0 + 0, ar1, n1 + 0, m1)
Proof step 6:
∀ 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

indvar: m0

Additional information

Used simplifier rules: nat freely generated by 0, +1;;