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Proof step 24 from proof orderedls-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
m1 + 1 < n, a < a0, ordered<(a0 ' + y), n ≤ 1 + # y, m1 < # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ y = [], (a0 ' + y)[m1] < (a ' + a0 ' + y)[n]

Rule argument

Additional information

Used simplifier rules: l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2