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Proof step 4 from proof ole-in (previous step)

b < a, ordered≤(a ' + (a0 + x0)), b ∈ a0 + x0, ∀ b, a. ¬(b < a ∧ ordered≤(a ' + x0) ∧ b ∈ x0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
b < a, ordered≤(a0 ' + x0), b ∈ x0, ∀ b0, a1. ¬ b0 < a1 ∨ ¬ ordered≤(a1 ' + x0) ∨ ¬ b0 ∈ x0 ⊦ a0 < a

Rule argument

Additional information

Used simplifier rules: le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);%cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%app-01: ⊦ ordered≤(x + y) → ordered≤(y);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)