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Proof step 4 from proof range-eq-ext (previous step)

∀ 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 < # ar0 ∧ n1 + m0 < # ar1 ∧ (∀ n. ¬ m0 < n → ar0[n0 + n] = ar1[n1 + n])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ range=(ar0, n0, ar1, n1, m + 1) ↔ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m)
Proof step 6:
∀ 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]) ⊦ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m0) ↔ n0 + m0 < # ar0 ∧ n1 + m0 < # ar1 ∧ (∀ n. ¬ m0 < n → ar0[n0 + n] = ar1[n1 + n])

Rule argument

spec: <toplevel> name: range-eq-rec seq: ⊦ range=(ar0, n0, ar1, n1, m + 1) ↔ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m) subst: [m, n1, ar1, n0, ar0] → [m0, n1, ar1, n0, ar0] flags: <all paths>

Additional information

Used terms: m0, n1, ar1, n0, ar0