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Proof step 4 from proof min-two (previous step)

a < b ⊦ (∅ ++ a ++ b).min = a, b < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
a ∈ s, ∀ b. b ∈ s → ¬ b < a ⊦ a = s.min

Rule argument

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

Additional information

Used simplifier rules: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ a ∈ ∅ ++ b ↔ a = b; (s)% ⊦ ¬ a < a; (s) Used terms: a, ∅ ++ a ++ b