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

n < # kar, no-dups-idx(kar, n) ⊦ no-dups-idx(kar[n, ki], n + 1) ↔ ¬(∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = ki.idx)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # kar, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ (∀ m, m0. m < # kar ∧ ¬ n < m ∧ m0 < # kar ∧ ¬ n < m0 ∧ m ≠ m0 → kar[n, ki][m].idx ≠ kar[n, ki][m0].idx) ↔ ¬(∃ ki0. ki0 ∈ keyset(kar, n) ∧ ki0.idx = ki.idx)

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 < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%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)% ⊦ # kar[n, ki] = # kar; (s)