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Proof step 31 from proof orderedls-all (previous step)

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
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)

Proof step 31 from proof orderedls-all (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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