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Proof step 5 from proof sorted-lt-h (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 6:
m2 = m +1, sorted(zbrar, n, m2 + 1), m1 ≤ m2, m0 ≤ 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

var: m2

Additional information

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