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Proof step 17 from proof orderedls-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
a < a0, (a0 ' + y)[m] < a0, ordered<(a0 ' + y), m ≤ # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ a < (a0 ' + y)[m]

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem% ⊦ 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)