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Proof step 19 from proof ordered-prefix (previous step)

x1 = a1 + y, ordered≤(a0 ' + y0), x1 ⊑ a0 ' + y0, (∃ x0. ordered≤(x0) ∧ x1 ⊑ x0) → ordered≤(x1) ⊦ x1 = [], a0 < a, ¬ a1 < a ∧ ordered≤(x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
ordered≤(a0 ' + y0), y ⊑ y0 ⊦ a0 < a, ordered≤(a0 ' + y), ∃ x. ordered≤(x) ∧ a0 ' + y ⊑ x

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ⊑ b ' + y ↔ a = b ∧ x ⊑ y; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)