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

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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