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Proof step 11 from proof no-dups-idx-rec (previous step)

n < # kar, n ≤ # kar, ∀ 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, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx, ∃ m. m < n ∧ key-index(key0, m2) = kar[m] ⊦ # keyset(kar, n) = 0, keyset(kar, n) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
key-index(key0, m2) = kar[m], m < # kar, n < # kar, m < n, m + 1 < # kar, ∀ 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, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ # keyset(kar, n) = 0, keyset(kar, n) = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)