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Proof step 3 from proof orderedls-all (previous step)

ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ ordered<(a ' + x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ x ≠ [] ∧ x.tail ≠ [] → (ordered<(x) ↔ x.head < x.tail.head ∧ ordered<(x.tail))
Proof step 5:
ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ a ' + x0 = [], (a ' + x0).tail = [], (a ' + x0).head < (a ' + x0).tail.head ∧ ordered<((a ' + x0).tail) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n])

Rule argument

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

Additional information

Used simplifier rules: ls-e: ⊦ ordered<([]);%ls-o: ⊦ ordered<(a ');% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s)% ⊦ # [] = 0; (s)% ⊦ x + [] = x; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m < n → n ≠ 0; (s) Used terms: a ' + x0