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Proof step 9 from proof orderedls-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a < a0, m < n, ordered<(a0 ' + y), n ≤ 1 + # y, m ≤ # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ (a ' + a0 ' + y)[m] < (a ' + a0 ' + y)[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)