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Proof step 3 from proof key-set-extract-index (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 4:
n0 < # kar, n0 ≤ # kar → (ki ∈ keyset(kar, n0) ↔ (∃ m. m < n0 ∧ ki = kar[m])) ⊦ ki = kar[n0] ∨ ki ∈ keyset(kar, n0) ↔ (∃ m. ¬ n0 < m ∧ ki = kar[m])

Rule argument

pos: left 2 terms: kar, ki abort: F discard: T computed: T

Additional information

Used terms: kar, ki