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Proof step 19 from proof orderedls-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
y[m1] < a0, a < a0, m1 < # y, ordered<(a0 ' + y), ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ y = [], a < y[m1]

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x)[n + 1] = x[n]; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)