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Proof step 15 from proof del1 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ x ≠ [] ∧ x.tail ≠ [] → (ordered<(x) ↔ x.head < x.tail.head ∧ ordered<(x.tail))
Proof step 17:
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

Rule argument

spec: <toplevel> name: ols-rec seq: ⊦ x ≠ [] ∧ x.tail ≠ [] → ( ordered<(x) ↔ x.head < x.tail.head ∧ ordered<(x.tail)) subst: [x] → [a0 ' + x0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ [] -1l a = []; (s)% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s) Used terms: a0 ' + x0