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Proof step 24 from proof ole-last (previous step)

x0 + a2 ' = a1 ' + y, y = y0 + a2 ', a1 < a2, a0 < a2, ordered≤(a1 ' + y), a1 ∈ x0, ∀ a, b. ordered≤(x0 + a ') → (ordered≤(x0 + a ' + b ') ↔ ¬ b < a) ⊦ a1 = a2, y = [], b < a2, a1 < a0, b < a1, b < a0, ordered≤((a0 ' + x0) + a2 ' + b ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
a0 < a2, a1 < a2, ordered≤(a1 ' + y0 + a2 '), ∀ a, b. ordered≤((a1 ' + y0) + a ') → (ordered≤((a1 ' + y0) + a ' + b ') ↔ ¬ b < a) ⊦ a1 = a2, b < a2, a1 < a0, b < a1, b < a0, ordered≤(a1 ' + y0 + a2 ' + b ')

Rule argument

Additional information

Used simplifier rules: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%app-01: ⊦ ordered≤(x + y) → ordered≤(y);%le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);% ⊦ a + x = a ' + x; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ x + a ' ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = z + y ↔ x = z; (s)