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Proof step 9 from proof key-set-extract-index (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m < # kar, m < n0, n0 < # kar, m + 1 < # kar, kar[m] ∈ keyset(kar, n0) ⊦ # keyset(kar, n0) = 0, keyset(kar, n0) = ∅, ∃ m0. ¬ n0 < m0 ∧ kar[m] = kar[m0]

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic[keyindex-set]% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%ne-01: ⊦ ki ∈ s → s ≠ ∅; (forward) from specification set-basic[keyindex-set]%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