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Proof step 2 from proof ordered-ex-divide-lels (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ ∃ y, z. [] = y + z ∧ (∀ b. b ∈ y → b < a ∨ b = a) ∧ (∀ b. b ∈ z → a < b)

Rule argument

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]);