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

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

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] → [n, m + m0, avol0, m, n, avol] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n, m + m0, avol0, m, n, avol