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Proof step 8 from proof ordered-ex-divide-lels (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
ordered≤(a0 ' + y + z), a < b, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → b < a ∨ b = a, b = a0 ∨ b ∈ y ⊦ b = a, ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → b < a ∨ b = a) ∧ (∀ b. b ∈ z0 → a < b)

Rule argument

Additional information

Used simplifier rules: sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list