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Proof step 29 from proof orderedls-all (previous step)

n = m + 1, m1 < # y, a < a0, m1 + 1 < n, ordered<(a0 ' + y), n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ n = 0, y = [], (a0 ' + y)[m1] < (a0 ' + y)[m]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
m1 < # y, m1 < m, a < a0, ordered<(a0 ' + y), m ≤ # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ y = [], (a0 ' + y)[m1] < (a0 ' + y)[m]

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 < m + n0 ↔ n < n0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)