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Proof step 11 from proof key-set-extract-index (previous step)

ki ∈ keyset(kar, n0) ↔ (∃ m. m < n0 ∧ ki = kar[m]), n0 < # kar, ∃ m. ¬ n0 < m ∧ ki = kar[m] ⊦ ki = kar[n0], ki ∈ keyset(kar, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n0 < # kar, m < # kar, m ≤ n0 ⊦ kar[m] = kar[n0], kar[m] ∈ keyset(kar, n0), ∃ m0. m0 < n0 ∧ kar[m] = kar[m0]

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)