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Proof step 6 from proof max-com (previous step)

∀ x1, x2. # x1 + # x2 < #(a + x3) + # x0 → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ max((a ' + x3) + x0) = max(x0 + a ' + x3)

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))
Proof step 8:
x3 + x0 ≠ [], ∀ x1, x2. # x1 + # x2 < #(a + x3) + # x0 → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (max(x3 + x0) < a ⊃ a;max(x3 + x0)) = max(x0 + a ' + x3)

Rule argument

spec: inst: name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x3 + x0, a] flags: rewrite, current

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%Nil: ⊦ [] + x = x; from specification list%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x3 + x0, a