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Proof step 3 from proof k-le-r (previous step)

y = a + y0 ⊦ ordered≤(y) ↔ # y ≤ 1 ∨ ¬ y0.head < a ∧ ordered≤(y0), y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ordered≤(a ' + y0) ↔ y0 = [] ∨ ¬ y0.head < a ∧ ordered≤(y0)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)