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Proof step 33 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 34:
m < n, n ≤ # y, m < # y, ∀ m, n. m < n ∧ ¬ 1 + # y < n → (a ' + a0 ' + y)[m] < (a ' + a0 ' + y)[n] ⊦ y = [], ordered<(a0 ' + y), (a0 ' + y)[m] < (a0 ' + y)[n]

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2