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Proof step 3 from proof min-ins-02 (previous step)

s.min < (s ++ a).min, s.min ∈ s, s.max ∈ s ⊦ # s = 0, s = ∅, s.max < s.min

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
a ∈ s, ∀ b. b ∈ s → ¬ b < a ⊦ a = s.min
Proof step 5:
s.min < (s ++ a).min, s.max ∈ s, s.min ∈ s ⊦ # s = 0, s = ∅, s.max < s.min, s.min ∈ s ++ a ∧ (∀ b. b ∈ s ++ a → ¬ b < s.min)

Rule argument

spec: <toplevel> name: min-q seq: a ∈ s, ∀ b. b ∈ s → ¬ b < a ⊦ a = s.min subst: [a, s] → [s.min, s ++ a] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ a ∈ s → a ∈ s ++ b; (s)% ⊦ ¬ a < a; (s) Used terms: s.min, s ++ a