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Proof step 3 from proof range-eq-extract (previous step)

n < n2, range=(ar0, n0, ar1, n1, n2) ⊦ ar0[n0 + n] = ar1[n1 + n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ range=(ar0, n0, ar1, n1, m) ↔ n0 + m ≤ # ar0 ∧ n1 + m ≤ # ar1 ∧ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n])
Proof step 5:
n < n2, n0 + n2 ≤ # ar0, n1 + n2 ≤ # ar1, ∀ n. n < n2 → ar0[n0 + n] = ar1[n1 + n] ⊦ ar0[n0 + n] = ar1[n1 + n]

Rule argument

spec: <toplevel> name: range-eq-ext seq: ⊦ range=(ar0, n0, ar1, n1, m) ↔ n0 + m ≤ # ar0 ∧ n1 + m ≤ # ar1 ∧ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n]) subst: [m, n1, ar1, n0, ar0] → [n2, n1, ar1, n0, ar0] flags: <all paths>

Additional information

Used terms: n2, n1, ar1, n0, ar0