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Proof step 18 from proof ole-01 (previous step)

ins≤(a, a2 ' + y0) = a1 ' + y, a0 < a, ∀ a3. ordered≤(ins≤(a3, a2 ' + y0)) ↔ ordered≤(a2 ' + y0) ⊦ ¬ a1 < a0 ∧ ordered≤(a1 ' + y) ↔ ¬ a2 < a0 ∧ ordered≤(a2 ' + y0), a2 < a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;%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);%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem% ⊦ ¬ a < a; (s)%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)