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Proof step 6 from proof min-ins-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ s → ¬ a < s.min

Rule argument

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

Additional information

Used simplifier rules: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws) Used terms: s ++ a, a