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Proof step 22 from proof orderedls-all (previous step)

m < n, a < a0, ordered<(a0 ' + y), m ≤ # y, n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ a ' + a0 ' + y = [], m = 0, (a ' + a0 ' + y).tail[m - 1] < (a ' + a0 ' + y)[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
a < a0, m < n, ordered<(a0 ' + y), n ≤ 1 + # y, m ≤ # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ m = 0, (a0 ' + y)[m - 1] < (a ' + a0 ' + y)[n]

Rule argument

Additional information

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