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Proof step 2 from proof no-dups-idx-mvlast (previous step)

kar[m] = ki, m < # kar, no-dups-idx(kar, m + 1) ⊦ no-dups-idx(kar[m0, ki], m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
kar[m0, kar[m]][m1].idx = kar[m0, kar[m]][m2].idx, m2 < m, m1 < m, m1 < # kar, m < # kar, m2 + 1 < # kar, m2 < # kar, m1 + 1 < # kar, ∀ m1, m2. m1 < # kar ∧ ¬ m < m1 ∧ m2 < # kar ∧ ¬ m < m2 ∧ m1 ≠ m2 → kar[m1].idx ≠ kar[m2].idx ⊦ m1 = m2

Rule argument

Additional information

Used simplifier rules: no-dups-idx: ⊦ no-dups-idx(kar, n) ↔ (∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx);% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # kar[n, ki] = # kar; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)