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Proof step 6 from proof sorted-all (previous step)

n + m3 + 1 < # zbrar, ∀ m, m0. m < m0 ∧ ¬ m3 + 1 < m0 → zbrar[n + m].key < zbrar[n + m0].key, ∀ n0, zbrar0. n0 + m3 < # zbrar0 ∧ (∀ m, m0. m < m0 ∧ ¬ m3 < m0 → zbrar0[n0 + m].key < zbrar0[n0 + m0].key) → sorted(zbrar0, n0, m3 + 1) ⊦ sorted(zbrar, n, 2 + m3)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ sorted(zbrar, n, m + 2) ↔ n < # zbrar ∧ zbrar[n].key < zbrar[n + 1].key ∧ sorted(zbrar, n + 1, m + 1)
Proof step 8:
n + m3 + 1 < # zbrar, ∀ n0, zbrar0. n0 + m3 < # zbrar0 ∧ (∀ m, m0. m < m0 ∧ ¬ m3 < m0 → zbrar0[n0 + m].key < zbrar0[n0 + m0].key) → sorted(zbrar0, n0, m3 + 1), ∀ m, m0. m < m0 ∧ ¬ m3 + 1 < m0 → zbrar[n + m].key < zbrar[n + m0].key ⊦ n < # zbrar ∧ zbrar[n].key < zbrar[n + 1].key ∧ sorted(zbrar, n + 1, m3 + 1)

Rule argument

spec: <toplevel> name: sorted-rec seq: ⊦ sorted(zbrar, n, m + 2) ↔ n < # zbrar ∧ zbrar[n].key < zbrar[n + 1].key ∧ sorted(zbrar, n + 1, m + 1) subst: [m, n, zbrar] → [m3, n, zbrar] flags: <all paths>

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic Used terms: m3, n, zbrar