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Proof step 4 from proof sorted-all (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 5:
m0 = m3 +1, n + m0 < # zbrar, ∀ n, zbrar. n + m0 ≤ # zbrar ∧ (∀ m2, m1. m2 < m1 ∧ m1 < m0 → zbrar[n + m2].key < zbrar[n + m1].key) → sorted(zbrar, n, m0), ∀ m, m1. m < m1 ∧ ¬ m0 < m1 → zbrar[n + m].key < zbrar[n + m1].key ⊦ sorted(zbrar, n, m0 + 1)

Rule argument

var: m0

Additional information

Used simplifier rules: sorted-one: ⊦ sorted(zbrar, n, 1);%nat freely generated by 0, +1;;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic