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Proof step 17 from proof max-rec (previous step)

i ≤ max(i ' + ints), max(ints) ≤ max(i ' + ints), ∀ i0. i0 ∈ ints → i0 ≤ max(i ' + ints), i ∈ ints → i ≤ max(ints), ∀ i. i ∈ ints → i ≤ max(ints) ⊦ ints = [], max(i ' + ints) = max(i, max(ints)), max(i ' + ints) ∈ ints

User applied rule cut formula

And obtained the following new goals

Proof step 18:
max(i ' + ints) ∈ i ' + ints, i ≤ max(i ' + ints), max(ints) ≤ max(i ' + ints), ∀ i0. i0 ∈ ints → i0 ≤ max(i ' + ints), i ∈ ints → i ≤ max(ints), ∀ i. i ∈ ints → i ≤ max(ints) ⊦ ints = [], max(i ' + ints) = max(i, max(ints)), max(i ' + ints) ∈ ints

Rule argument

fma: max(i ' + ints) ∈ i ' + ints

Additional information

Used simplifier rules: max-in-01: ⊦ max(ints) ∈ ints ↔ ints ≠ [];%e-01: ⊦ i ' + ints ≠ []; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista] Used formulas: max(i ' + ints) ∈ i ' + ints