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Proof step 13 from proof sorted-lt-h (previous step)

zbrar[n].key < zbrar[n + 1].key, n < # zbrar, sorted(zbrar, n + 1, m + 1), m0 ≤ m + 1, m4 ≤ m, ∀ m2, m3, zbrar0, n0. ¬ m < m2 ∧ ¬ m < m3 ∧ sorted(zbrar0, n0, m + 1) → (zbrar0[n0 + m2].key < zbrar0[n0 + m3].key ↔ m2 < m3) ⊦ zbrar[n + m0].key < zbrar[n + m4 + 1].key ↔ ¬ m4 < m0

User applied rule constructor cut

And obtained the following new goals

Proof step 14:
m0 = 0, n < # zbrar, zbrar[n].key < zbrar[n + 1].key, sorted(zbrar, n + 1, m + 1), m4 ≤ m, m0 ≤ m + 1, ∀ m2, m3, zbrar0, n0. ¬ m < m2 ∧ ¬ m < m3 ∧ sorted(zbrar0, n0, m + 1) → (zbrar0[n0 + m2].key < zbrar0[n0 + m3].key ↔ m2 < m3) ⊦ zbrar[n + m0].key < zbrar[n + m4 + 1].key ↔ ¬ m4 < m0
Proof step 16:
m0 = m1 +1, n < # zbrar, zbrar[n].key < zbrar[n + 1].key, sorted(zbrar, n + 1, m + 1), m4 ≤ m, m0 ≤ m + 1, ∀ m2, m3, zbrar0, n0. ¬ m < m2 ∧ ¬ m < m3 ∧ sorted(zbrar0, n0, m + 1) → (zbrar0[n0 + m2].key < zbrar0[n0 + m3].key ↔ m2 < m3) ⊦ zbrar[n + m0].key < zbrar[n + m4 + 1].key ↔ ¬ m4 < m0

Rule argument

var: m0

Additional information

Used simplifier rules: nat freely generated by 0, +1;;