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Proof step 21 from proof no-dups-keyset-size (previous step)

#(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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ ¬ ki ∈ s → #(s ++ ki) = # s + 1

Rule argument

spec: set-basic, inst: keyindex-set name: Size-insert seq: ⊦ ¬ ki ∈ s → #(s ++ ki) = # s + 1 subst: [ki, s] → [kar[m0], keyset(kar, m0)] flags: <all paths>

Additional information

Used simplifier rules: keyset-size-bound0: m < # kar ⊦ # keyset(kar, m) = m + 1 ↔ false;% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ki ∈ s → s ++ ki = s; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: kar[m0], keyset(kar, m0)