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Proof step 10 from proof del1-01 (previous step)

ordered≤((a0 ' + x0).tail), ∀ a1. ordered≤(x0 -1l a1) ⊦ a0 ' + x0 = [], (a0 ' + x0).tail = [], a = a0, (a0 ' + x0).tail.head < (a0 ' + x0).head, ordered≤(a0 ' + (x0 -1l a))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
ordered≤(x0), ∀ a1. ordered≤(x0 -1l a1) ⊦ a = a0, x0 = [], x0.head < a0, ordered≤(a0 ' + (x0 -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)