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Proof step 2 from proof no-dups-shorter (previous step)

no-dups(kar, m + 1) ⊦ no-dups(kar, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: no-dups: ⊦ no-dups(kar, n) ↔ (∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0] ≠ kar[m1]);% ⊦ 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)