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Proof step 6 from proof key-set-extract-index (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
ki ∈ keyset(kar, n0) ↔ (∃ m. m < n0 ∧ ki = kar[m]), ki = kar[n0], n0 < # kar ⊦ ∃ m. ¬ n0 < m ∧ ki = kar[m]
Proof step 9:
ki ∈ keyset(kar, n0) ↔ (∃ m. m < n0 ∧ ki = kar[m]), n0 < # kar, ki ∈ keyset(kar, n0) ⊦ ∃ m. ¬ n0 < m ∧ ki = kar[m]

Rule argument

pos: left 3

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