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Proof step 10 from proof keep-sorted (previous step)

zbr.key = zbrar[n0].key, n0 < # zbrar, ∀ zbr0, zbrar0, n1, n2. zbr0.key = zbrar0[n1].key ∧ n1 < # zbrar0 → (sorted(zbrar0[n1, zbr0], n2, m + 1) ↔ sorted(zbrar0, n2, m + 1)) ⊦ n < # zbrar ∧ zbrar[n0, zbr][n].key < zbrar[n0, zbr][n + 1].key ∧ sorted(zbrar[n0, zbr], n + 1, m + 1) ↔ sorted(zbrar, n, 2 + m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ sorted(zbrar, n, m + 2) ↔ n < # zbrar ∧ zbrar[n].key < zbrar[n + 1].key ∧ sorted(zbrar, n + 1, m + 1)
Proof step 12:
zbr.key = zbrar[n0].key, n0 < # zbrar, ∀ zbr0, zbrar0, n1, n2. zbr0.key = zbrar0[n1].key ∧ n1 < # zbrar0 → (sorted(zbrar0[n1, zbr0], n2, m + 1) ↔ sorted(zbrar0, n2, m + 1)) ⊦ n < # zbrar ∧ zbrar[n0, zbr][n].key < zbrar[n0, zbr][n + 1].key ∧ sorted(zbrar[n0, zbr], n + 1, m + 1) ↔ n < # zbrar ∧ zbrar[n].key < zbrar[n + 1].key ∧ sorted(zbrar, n + 1, m + 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] → [m, n, zbrar] flags: <all paths>

Additional information

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