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Proof step 4 from proof max-ifins (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
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; (s)%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem% ⊦ ¬ a < a; (s) Used terms: s.max, s ++ a