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Proof step 20 from proof no-dups-keyset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
#(keyset(kar, m0) ++ kar[m0]) = m0 + 1, kar[m1] = kar[m], kar[m3] = kar[m2], m0 < # kar, m1 < # kar, m < # kar, m2 < m0, m3 < m0, m3 < # kar, m2 + 1 < # kar, m2 < # kar, m3 + 1 < # kar, m1 ≤ m0, m ≤ m0 ⊦ # keyset(kar, m0) = m0, m1 = m, m3 = m2

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic