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Proof step 13 from proof no-dups-keyset-size (previous step)

# keyset(kar, m0) = m0, m0 < # kar, kar[m0] ∈ keyset(kar, m0), ∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2] ≠ kar[m1], ∀ m, m1. m < # kar ∧ ¬ m0 < m ∧ m1 < # kar ∧ ¬ m0 < m1 ∧ m ≠ m1 → kar[m] ≠ kar[m1] ⊦ #(keyset(kar, m0) ++ kar[m0]) = m0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# keyset(kar, m0) = m0, m0 < # kar, kar[m0] ∈ keyset(kar, m0), ∀ m, m1. m < # kar ∧ ¬ m0 < m ∧ m1 < # kar ∧ ¬ m0 < m1 ∧ m ≠ m1 → kar[m] ≠ kar[m1], ∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2] ≠ kar[m1] ⊦ keyset(kar, m0) = ∅, m0 = 0

Rule argument

Additional information

Used simplifier rules: keyset-size-bound0: m < # kar ⊦ # keyset(kar, m) = m + 1 ↔ false;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic[keyindex-set]%ne-01: ⊦ ki ∈ s → s ≠ ∅; (forward) from specification set-basic[keyindex-set]%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (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 → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ki ∈ s → s ++ ki = s; (s)