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Proof step 2 from proof rangeeq-lebs-5 (previous step)

range=(avol, 0, avol0, 0, m + m0), n ≤ m0 ⊦ lebs(avol, m, n) = lebs(avol0, m, n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ m ≤ n0 ∧ n0 + n1 ≤ m + m2 ∧ range=(avol0, m, avol1, m, m2) → lebs(avol0, n0, n1) = lebs(avol1, n0, n1)

Rule argument

spec: persistence-io-inv name: lebs-range-eq seq: ⊦ m ≤ n0 ∧ n0 + n1 ≤ m + m2 ∧ range=(avol0, m, avol1, m, m2) → lebs(avol0, n0, n1) = lebs(avol1, n0, n1) subst: [m, m2, avol1, n1, n0, avol0] → [0, m + m0, avol0, n, m, avol] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ 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 + n ≤ m + n0 ↔ n ≤ n0; (s) Used terms: 0, m + m0, avol0, n, m, avol