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Proof step 28 from proof no-dups-idx-rec (previous step)

kar[m] = key-index(key1, m3), m + 1 < # kar, m < # kar, n < # kar, m < n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m = n, ∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m3

User applied rule exists right

And obtained the following new goals

Proof step 29:
kar[m] = key-index(key1, m3), m < n, n < # kar, m < # kar, m + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m = n, key-index(key1, m3) ∈ keyset(kar, n) ∧ key-index(key1, m3).idx = m3

Rule argument

pos: right 2 terms: key-index(key1, m3) abort: F discard: T computed: F

Additional information

Used terms: key-index(key1, m3)