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Proof step 4 from proof sorted-lt-h (previous step)

m1 < m2 +1, m0 < m2 +1, sorted(zbrar, n, m2 +1), ∀ m0, m1, zbrar, n. m0 < m2 ∧ m1 < m2 ∧ sorted(zbrar, n, m2) → (zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1) ⊦ zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
sorted(zbrar, n, m2 + 1), m0 ≤ m2, m1 ≤ m2, ∀ m0, m1, zbrar, n. m0 < m2 ∧ m1 < m2 ∧ sorted(zbrar, n, m2) → (zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1) ⊦ zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)