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Proof step 15 from proof del1-01 (previous step)

ordered≤(a2 ' + y), ∀ a1. ordered≤((a2 ' + y) -1l a1) ⊦ a0 ' + ((a2 ' + y) -1l a) = [], (a0 ' + ((a2 ' + y) -1l a)).tail = [], a = a0, a2 < a0, ¬ (a0 ' + ((a2 ' + y) -1l a)).tail.head < (a0 ' + ((a2 ' + y) -1l a)).head ∧ ordered≤((a0 ' + ((a2 ' + y) -1l a)).tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
ordered≤(a2 ' + y), ∀ a1. ordered≤((a2 ' + y) -1l a1) ⊦ (a2 ' + y) -1l a = [], a = a0, a2 < a0, ¬ ((a2 ' + y) -1l a).head < a0 ∧ ordered≤((a2 ' + y) -1l a)

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)