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Proof step 5 from proof ordered-ex-divide-lels (previous step)

ordered≤(a0 ' + x0), ∀ a1. ∃ y, z. x0 = y + z ∧ (∀ b. b ∈ y → b < a1 ∨ b = a1) ∧ (∀ b. b ∈ z → a1 < b) ⊦ ∃ y, z. a0 ' + x0 = y + z ∧ (∀ b. b ∈ y → b < a ∨ b = a) ∧ (∀ b. b ∈ z → a < b)

User applied rule all left

And obtained the following new goals

Proof step 6:
ordered≤(a0 ' + x0), ∃ y, z. x0 = y + z ∧ (∀ b. b ∈ y → b < a ∨ b = a) ∧ (∀ b. b ∈ z → a < b) ⊦ ∃ y, z. a0 ' + x0 = y + z ∧ (∀ b. b ∈ y → b < a ∨ b = a) ∧ (∀ b. b ∈ z → a < b)

Rule argument

pos: left 2 terms: a abort: F discard: T computed: F

Additional information

Used terms: a