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Proof step 6 from proof ordered-ex-divide-lels (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
ordered≤(a0 ' + y + z), ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → b < a ∨ 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: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%app-01: ⊦ ordered≤(x + y) → ordered≤(y);%a: ⊦ (x + y) + z = x + y + z; from specification list