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Proof step 9 from proof app (previous step)

x0 = a0 + y0, ordered≤((a ' + x0) + y), (∃ y. ordered≤(x0 + y)) → ordered≤(x0) ⊦ x0 = [], ¬ a0 < a ∧ ordered≤(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ordered≤(a0 ' + y0 + y) ⊦ a0 < a, ordered≤(a0 ' + y0), ∃ y1. ordered≤((a0 ' + y0) + y1)

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 ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)