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Proof step 9 from proof butmin-ins (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ a ∈ s ∧ a ≠ b → (s -- b) ++ a = s -- b
Proof step 11:
s.min < a, s.max ∈ s, s.min ∈ s ⊦ # s = 0, (s ++ a) -- s.min = (s -- s.min) ++ a, s = ∅, a ∈ s ∧ a ≠ s.min

Rule argument

spec: set-union name: del-addeq-03 seq: ⊦ a ∈ s ∧ a ≠ b → (s -- b) ++ a = s -- b subst: [a, b, s] → [a, s.min, s] flags: <all paths>

Additional information

Used simplifier rules: ⊦ a ∈ s → s ++ a = s; (s)% ⊦ a ∈ s ∧ a ∈ s0 → (s -- a = s0 -- a ↔ s = s0); (s) Used terms: a, s.min, s