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

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

User applied rule insert rewrite lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]);%le-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)% ⊦ 0 < n ↔ n ≠ 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ a '[0] = a; (s)% ⊦ ¬ a < a; (s) Used terms: a ' + x0