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Proof step 7 from proof orderedle-all (previous step)

x0 = a0 + y, ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ n < # x0 → ¬ x0[n] < x0[m]) ⊦ ¬ a0 < a ∧ ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ ¬ # x0 < n → ¬ (a ' + x0)[n] < (a ' + x0)[m]), x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
ordered≤(a0 ' + y) ↔ (∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m]) ⊦ ¬ a0 < a ∧ ordered≤(a0 ' + y) ↔ (∀ m, n. m ≤ n ∧ ¬ 1 + # y < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m])

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%c: ⊦ a + x = a ' + x; from specification list