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Proof step 4 from proof ordered-ex-divide-lels (previous step)

∀ a. ordered≤(x0) → (∃ y, z. x0 = y + z ∧ (∀ b. b ∈ y → b < a ∨ b = a) ∧ (∀ b. b ∈ z → a < b)), ordered≤(a0 + x0) ⊦ ∃ y, z. a0 + x0 = 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 5:
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)

Rule argument

Additional information

Used simplifier rules: app-01: ⊦ ordered≤(x + y) → ordered≤(y);%cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list