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Proof step 6 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, key-index(key, m2)][m].idx ≠ kar[n, key-index(key, m2)][m0].idx, ∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m2 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < # kar, ki ∈ keyset(kar, n), ∀ 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, key-index(key, ki.idx)][m].idx ≠ kar[n, key-index(key, ki.idx)][m0].idx ⊦ # keyset(kar, n) = 0, keyset(kar, n) = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic[keyindex-set]%ne-01: ⊦ ki ∈ s → s ≠ ∅; (forward) from specification set-basic[keyindex-set]