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Proof step 18 from proof no-dups-keyset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
#(keyset(kar, m0) ++ kar[m0]) = m0 + 1, (∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2] ≠ kar[m1]) ↔ # keyset(kar, m0) = m0, kar[m1] = kar[m], m < # kar, m1 < # kar, m0 < # kar, m ≤ m0, m1 ≤ m0 ⊦ m1 = m

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≤ n ↔ m < n; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)