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Proof step 22 from proof orderedle-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
m1 < # y, y[m1] < a, ordered≤(a0 ' + y), ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ y = [], a0 < a

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%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