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Proof step 11 from proof ordered-ex-divide-lels (previous step)

ordered≤(a0 ' + y + z), a < a0, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → b < a ∨ b = a ⊦ a0 = a, ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → b < a ∨ b = a) ∧ (∀ b. b ∈ z0 → a < b)

User applied rule exists right

And obtained the following new goals

Proof step 12:
ordered≤(a0 ' + y + z), a < a0, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → b < a ∨ b = a ⊦ a0 = a, a0 ' + y + z = [] + a0 ' + y + z ∧ (∀ b. b ∈ [] → b < a ∨ b = a) ∧ (∀ b. b ∈ a0 ' + y + z → a < b)

Rule argument

pos: right 2 terms: [], a0 ' + y + z abort: F discard: T computed: F

Additional information

Used terms: [], a0 ' + y + z