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

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

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list