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Proof step 10 from proof ordered-max (previous step)

max(x0) = x0.last, x0 = a0 + y, x0.last < a, ordered≤(x0) ⊦ a = x0.last, x0 = [], a0 < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
max(a0 ' + y) = (a0 ' + y).last, (a0 ' + y).last < a, ordered≤(a0 ' + y) ⊦ a = (a0 ' + y).last, a0 < a

Rule argument

Additional information

Used simplifier rules: lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)