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Proof step 17 from proof del1 (previous step)

x0 -1l a = a2 ' + y, (a0 ' + x0).head < (a0 ' + x0).tail.head, ordered<((a0 ' + x0).tail), ordered<(x0 -1l a), a2 ∈ x0, a2 ∈ x0 -1l a ⊦ a0 ' + x0 = [], x0 -1l a = [], (a0 ' + x0).tail = [], a = a0, a0 < a2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
x0 -1l a = a2 ' + y, a0 < x0.head, a2 < x0.head, ordered<(x0 -1l a), ordered<(x0), a2 ∈ x0 -1l a, a2 ∈ x0 ⊦ x0 -1l a = [], a = a0, x0 = [], a0 < a2

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem% ⊦ (a ' + x).head = a; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)