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Proof step 27 from proof orderedle-all (previous step)

m = m2 + 1, m ≤ 1 + # y, m0 ≤ # y, ordered≤(a0 ' + y), m ≤ m0 + 1, (a0 ' + y)[m0] < (a0 ' + y)[m2], ∀ m1, n. m1 ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m1] ⊦ m = 0, a0 < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
(a0 ' + y)[m0] < (a0 ' + y)[m2], m0 ≤ # y, ordered≤(a0 ' + y), m2 ≤ m0, m2 ≤ # y, ∀ m1, n. m1 ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m1] ⊦ a0 < a

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%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