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Proof step 3 from proof max-del-01 (previous step)

s.min ∈ s, s.max ∈ s ⊦ # s = 0, (s -- a).max = s.max, a = s.max, 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 → ¬ a < b ⊦ s.max = a

Rule argument

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

Additional information

Used simplifier rules: max-gr: ⊦ s.max < a ∧ s ≠ ∅ → ¬ a ∈ s;% ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; (ws) Used terms: s.max, s -- a