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Proof step 17 from proof del1-01-02 (previous step)

a < a0, ordered≤((a0 ' + x0).tail), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a0 ' + x0 = [], (a0 ' + x0).tail = [], a = a0, ins≤(a, x0) = a ' + x0, (a0 ' + x0).tail.head < (a0 ' + x0).head

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
a < a0, ordered≤(x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = a0, ins≤(a, x0) = a ' + x0, x0 = [], x0.head < a0

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)