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Proof step 19 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, kar[m1] = kar[m], m < # kar, m1 < # kar, m0 < # kar, m ≤ m0, m1 ≤ m0 ⊦ m1 = m

User applied rule case distinction

And obtained the following new goals

Proof step 20:
#(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]
Proof step 23:
#(keyset(kar, m0) ++ kar[m0]) = m0 + 1, # keyset(kar, m0) = m0, kar[m1] = kar[m], m0 < # kar, m1 < # kar, m < # kar, m1 ≤ m0, m ≤ m0, ∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2] ≠ kar[m1] ⊦ m1 = m

Rule argument

pos: left 2

Proof step 19 from proof no-dups-keyset-size (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information