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Proof step 9 from proof min-delmin (previous step)

s.max = s.min, s = ∅ ++ s.min, s.min ∈ s ⊦ # s = 1, # s = 0, s = ∅, s.min < (s -- s.min).min, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 10:
⊦ ¬ a ∈ s → #(s ++ a) = # s + 1
Proof step 11:
s.max = s.min, s = ∅ ++ s.min, s.min ∈ s, ¬ s.min ∈ ∅ → #(∅ ++ s.min) = # ∅ + 1 ⊦ # s = 0, # s = 1, s = ∅, s.min < (s -- s.min).min, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

Rule argument

spec: set-basic name: Size-insert seq: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1 subst: [a, s] → [s.min, ∅]

Additional information

Used terms: s.min, ∅