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Proof step 8 from proof butmin-ins (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ s ∧ a ≠ b → (s -- b) ++ a = s -- b
Proof step 10:
¬(a ∈ s ∧ a ≠ s.min), s.min < a, s.min ∈ s, s.max ∈ s ⊦ s = ∅, (s ++ a) -- s.min = (s -- s.min) ++ a, # s = 0

Rule argument

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

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s; from specification set-basic%del-eq: ⊦ a ∈ s ∧ a ∈ s0 → (s -- a = s0 -- a ↔ s = s0); from specification set-union Used terms: a, s.min, s