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Proof step 15 from proof max-com (previous step)

max(x3 + a0 ' + x) = max((a0 ' + x) + x3), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (max((a0 ' + x) + x3) < a ⊃ a;max((a0 ' + x) + x3)) = (max(x + a ' + x3) < a0 ⊃ a0;max(x + a ' + x3))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 17:
x + x3 = [], max(x3 + a0 ' + x) = max((a0 ' + x) + x3), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (max((a0 ' + x) + x3) < a ⊃ a;max((a0 ' + x) + x3)) = (max(x + a ' + x3) < a0 ⊃ a0;max(x + a ' + x3))
Proof step 20:
max(x3 + a0 ' + x) = (max(x + x3) < a0 ⊃ a0;max(x + x3)), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ x + x3 = [], ((max(x + x3) < a0 ⊃ a0;max(x + x3)) < a ⊃ a;(max(x + x3) < a0 ⊃ a0;max(x + x3))) = (max(x + a ' + x3) < a0 ⊃ a0;max(x + a ' + x3))

Rule argument

spec: <toplevel> name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x + x3, a0] flags: <all paths>

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x + x3, a0