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Proof step 5 from proof ordered-max (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ordered≤(y) ↔ # y ≤ 1 ∨ ¬ y.rest.first < y.first ∧ ordered≤(y.rest)
Proof step 7:
x0.last < a, max(x0) = x0.last, #(a ' + x0) ≤ 1 ∨ ¬ (a ' + x0).rest.first < (a ' + x0).first ∧ ordered≤((a ' + x0).rest) ⊦ a = x0.last, x0 = []

Rule argument

spec: olist inst: name: k-le-r seq: ⊦ ordered≤(y) ↔ # y ≤ 1 ∨ ¬ y.rest.first < y.first ∧ ordered≤(y.rest) subst: [y] → [a ' + x0] flags: rewrite

Additional information

Used terms: a ' + x0