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Proof step 3 from proof sorted-lt (previous step)

m0 < m, m1 < m, sorted(zbrar, 0, m) ⊦ zbrar[m0].key < zbrar[m1].key ↔ m0 < m1

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ m0 < m ∧ m1 < m ∧ sorted(zbrar, n, m) → (zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1)

Rule argument

spec: <toplevel> name: sorted-lt-h seq: ⊦ m0 < m ∧ m1 < m ∧ sorted(zbrar, n, m) → (zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1) subst: [m0, m, m1, zbrar, n] → [m0, m, m1, zbrar, 0]

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m0, m, m1, zbrar, 0