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

⊦ range=(ar0, n0, ar1, n1, m) ↔ n0 + m ≤ # ar0 ∧ n1 + m ≤ # ar1 ∧ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n])

User applied rule structural induction

And obtained the following new goals

Proof step 2:
⊦ range=(ar0, n0, ar1, n1, 0) ↔ n0 ≤ # ar0 ∧ n1 ≤ # ar1
Proof step 3:
∀ ar0, n0, ar1, n1. range=(ar0, n0, ar1, n1, m0) ↔ n0 + m0 ≤ # ar0 ∧ n1 + m0 ≤ # ar1 ∧ (∀ n. n < m0 → ar0[n0 + n] = ar1[n1 + n]) ⊦ range=(ar0, n0, ar1, n1, m0 +1) ↔ n0 + m0 +1 ≤ # ar0 ∧ n1 + m0 +1 ≤ # ar1 ∧ (∀ n. n < m0 +1 → ar0[n0 + n] = ar1[n1 + n])

Rule argument

indvar: 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% ⊦ ¬ m < 0; (s)