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Proof step 4 from proof no-dups-keyset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 < # kar, ∀ kar. m0 ≤ # kar → ((∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2] ≠ kar[m1]) ↔ # keyset(kar, m0) = m0) ⊦ (∀ m, m1. m < # kar ∧ ¬ m0 < m ∧ m1 < # kar ∧ ¬ m0 < m1 ∧ m ≠ m1 → kar[m] ≠ kar[m1]) ↔ #(keyset(kar, m0) ++ kar[m0]) = m0 + 1

Rule argument

Additional information

Used simplifier rules: keyset-rec: ⊦ n < # kar → keyset(kar, n + 1) = keyset(kar, n) ++ kar[n];% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (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)% ⊦ m +1 = m + 1; (s)