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Proof step 7 from proof no-dups-idx-remlast (previous step)

kar[0, kar[m]][m1] = key-index(key0, m2), kar[0, kar[m]][m0] = key-index(key, m2), m1 = 0, m1 < m, m0 < m, m0 < # kar, m < # kar, m1 + 1 < # kar, m1 < # kar, m0 + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ ¬ m < m0 ∧ m1 < # kar ∧ ¬ m < m1 ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m0 = m1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
kar[m0] = key-index(key, m2), kar[m] = key-index(key0, m2), m < # kar, m0 < # kar, m0 < m, m0 + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ ¬ m < m0 ∧ m1 < # kar ∧ ¬ m < m1 ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ # kar = 1, m0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m < # kar → kar[m, ki][m] = ki; (s)% ⊦ m ≠ n → kar[n, ki][m] = kar[m]; (s)