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Proof step 3 from proof ordered-max (previous step)

ordered≤(x0) → max(x0) = x0.last ∨ x0 = [], ordered≤(a + x0) ⊦ max(a + x0) = (a + x0).last, a + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
max(x0) = x0.last, ordered≤(a ' + x0) ⊦ x0 = [], (x0.last < a ⊃ a;x0.last) = x0.last

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%max-rec: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x));%l-02: ⊦ x ≠ [] → (a ' + x).last = x.last; from specification list-last%l: ⊦ a '.last = a; from specification list-last%app: ⊦ x + [] = x; from specification list%cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x); from specification olist%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list