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

x0 = a0 + y, ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ a < a0 ∧ ordered<(x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n]), 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)[m] < (a0 ' + y)[n]) ⊦ a < a0 ∧ ordered<(a0 ' + y) ↔ (∀ m, n. m < n ∧ ¬ 1 + # y < n → (a ' + a0 ' + y)[m] < (a ' + a0 ' + y)[n])

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