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Proof step 5 from proof orderedls-all (previous step)

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ a < x0.head ∧ ordered<(x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n]), x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).head = a; (s)