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Proof step 2 from proof k-ls-r (previous step)

y = a + y0 ∧ y ≠ [] ⊦ ordered<(y) ↔ # y ≤ 1 ∨ a < y0.first ∧ ordered<(y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list