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Proof step 7 from proof app (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
ordered≤((a ' + x0) + y), (∃ y. ordered≤(x0 + y)) → ordered≤(x0) ⊦ x0 = [], ¬ x0.head < a ∧ ordered≤(x0)

Rule argument

Additional information

Used simplifier rules: ⊦ (a ' + x).head = a; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)