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Proof step 1 from proof sorted-all (conclusion)

n + m ≤ # zbrar ⊦ sorted(zbrar, n, m) ↔ (∀ m0, m1. m0 < m1 ∧ m1 < m → zbrar[n + m0].key < zbrar[n + m1].key)

User applied rule case distinction

And obtained the following new goals

Proof step 2:
n + m ≤ # zbrar, ∀ m0, m1. m0 < m1 ∧ m1 < m → zbrar[n + m0].key < zbrar[n + m1].key ⊦ sorted(zbrar, n, m)

Rule argument

pos: right 1

Additional information

Used simplifier rules: sorted-lt-h: ⊦ m0 < m ∧ m1 < m ∧ sorted(zbrar, n, m) → (zbrar[n + m0].key < zbrar[n + m1].key ↔ m0 < m1);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic