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Proof step 14 from proof orderedle-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
ordered≤(a0 ' + y), m ≤ m0 + 1, (a0 ' + y)[m0] < (a ' + a0 ' + y)[m], m ≤ 1 + # y, m0 ≤ # y, ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ a0 < a

Rule argument

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; 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