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Proof step 3 from proof ordered-ex-divide-lels (previous step)

⊦ ∃ y, z. [] = y + z ∧ (∀ b. b ∈ y → b < a ∨ b = a) ∧ (∀ b. b ∈ z → a < b)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], [] abort: F discard: T computed: T

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-empty: ⊦ ¬ a ∈ []; from specification list Used terms: [], []