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

m = m1 +1, ordered<(a0 ' + y), a < a0, m ≤ # y, (a0 ' + y)[m] < a0, ∀ 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

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%a: ⊦ (x + y) + z = x + y + z; from specification list%get-succ: ⊦ (a ' + x)[n + 1] = x[n]; from specification list-del%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2