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Proof step 3 from proof app (previous step)

∀ y. ordered≤(x0 + y) → ordered≤(x0), ordered≤((a + x0) + y) ⊦ ordered≤(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ordered≤((a ' + x0) + y), (∃ y. ordered≤(x0 + y)) → ordered≤(x0) ⊦ ordered≤(a ' + x0)

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%c: ⊦ a + x = a ' + x; from specification list