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Proof step 6 from proof sort-max-last (previous step)

a < max(x1 + x0), max((a0 ' + x1) + x0) < max(x1 + x0) ⊦ ins≤(a0, sort(x1 + a ' + x0)) = ins≤(a0, sort(x1 + x0)) + a ', a < max((a0 ' + x1) + x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

Rule argument

spec: inst: name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x1 + x0, a0] flags: rewrite, current

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem%irreflexivity: ⊦ ¬ a < a; from specification oelem%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%Nil: ⊦ [] + x = x; from specification list%app: ⊦ x + [] = x; from specification list%one: ⊦ sort(a ') = a '; from specification olist-sort%empty: ⊦ sort([]) = []; from specification olist-sort%ins-e: ⊦ ins≤(a, []) = a '; from specification olist%ins-one-01: ⊦ ¬ b < a → ins≤(a, b ') = a ' + b '; from specification olist%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x1 + x0, a0