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Proof step 3 from proof no-dups-idx-rec (previous step)

n < # kar, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ (∀ m, m0. m < # kar ∧ ¬ n < m ∧ m0 < # kar ∧ ¬ n < m0 ∧ m ≠ m0 → kar[n, ki][m].idx ≠ kar[n, ki][m0].idx) ↔ ¬(∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = ki.idx)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
ki = key-index(key, m2), n < # kar, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ (∀ m, m0. m < # kar ∧ ¬ n < m ∧ m0 < # kar ∧ ¬ n < m0 ∧ m ≠ m0 → kar[n, ki][m].idx ≠ kar[n, ki][m0].idx) ↔ ¬(∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m2)

Rule argument

spec: key-index name: elim seq: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m) subst: [ki, key, m] → [ki, key, m2]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: ki, key, m2