Proof step 32 from proof orderedls-all (previous step)

User applied rule case distinction

And obtained the following new goals

• Proof step 33:

  ∀ m, n. m < n ∧ ¬ 1 + # y < n → (a ' + a0 ' + y)[m] < (a ' + a0 ' + y)[n] ⊦ ordered<(a0 ' + y), ∀ m, n. m < n ∧ n < 1 + # y → (a0 ' + y)[m] < (a0 ' + y)[n], a < a0 ∧ ordered<(a0 ' + y)

• Proof step 35:

  ordered<(a0 ' + y), ∀ m, n. m < n ∧ n < 1 + # y → (a0 ' + y)[m] < (a0 ' + y)[n], ∀ m, n. m < n ∧ ¬ 1 + # y < n → (a ' + a0 ' + y)[m] < (a ' + a0 ' + y)[n] ⊦ a < a0 ∧ ordered<(a0 ' + y)

Rule argument

Additional information