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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
n ≤ 1 + # y, ordered<(a0 ' + y), a < a0, m < n, 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: c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list