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Proof step 26 from proof no-dups-idx-rec (previous step)

kar[m] = key-index(key1, m3), kar[n, key-index(key, m2)][m0] = key-index(key0, m3), m < n, n < # kar, m < # kar, m0 < # kar, m + 1 < # kar, m0 ≤ n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m = n, m0 = m, ∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m2

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 27:
kar[n, key-index(key, m2)][m0] = key-index(key0, m3), kar[m] = key-index(key1, m3), m0 = n, m + 1 < # kar, m0 < # kar, m < # kar, n < # kar, m < n, m0 ≤ n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m = n, m0 = m, ∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m2
Proof step 34:
kar[n, key-index(key, m2)][m0] = key-index(key0, m3), kar[m] = key-index(key1, m3), m + 1 < # kar, m0 < # kar, m < # kar, n < # kar, m < n, m0 ≤ n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m = n, m0 = m, m0 = n, ∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = m2

Rule argument

fma: m0 = n

Additional information

Used formulas: m0 = n