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Proof step 3 from proof ordered-prefix (previous step)

∀ x0. x1 ⊑ x0 ∧ ordered≤(x0) → ordered≤(x1), a + x1 ⊑ x0, ordered≤(x0) ⊦ ordered≤(a + x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ordered≤(x0), a ' + x1 ⊑ x0, (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1) ⊦ ordered≤(a ' + x1)

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list