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Proof step 11 from proof no-dups-keyset-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ ¬ ki ∈ s → #(s ++ ki) = # s + 1
Proof step 13:
# keyset(kar, m0) = m0, m0 < # kar, kar[m0] ∈ keyset(kar, m0), ∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2] ≠ kar[m1], ∀ 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

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: ⊦ 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 Used terms: kar[m0], keyset(kar, m0)