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Proof step 4 from proof nodups-interval-size (previous step)

j1 < n→i(# ints) + j0, max(ints) ∈ ints → j0 < max(ints) ∧ max(ints) ≤ j1, ∀ i. i ∈ ints → j0 < i ∧ i ≤ j1, ∀ ints0, j0, j1. # ints0 < # ints → ¬ dups(ints0) ∧ (∀ i. i ∈ ints0 → j0 < i ∧ i ≤ j1) ∧ ints0 ≠ [] → n→i(# ints0) ≤ j1 - j0 ⊦ ints = [], dups(ints)

User applied rule cut formula

And obtained the following new goals

Proof step 5:
max(ints) ∈ ints, j1 < n→i(# ints) + j0, max(ints) ∈ ints → j0 < max(ints) ∧ max(ints) ≤ j1, ∀ i. i ∈ ints → j0 < i ∧ i ≤ j1, ∀ ints0, j0, j1. # ints0 < # ints → ¬ dups(ints0) ∧ (∀ i. i ∈ ints0 → j0 < i ∧ i ≤ j1) ∧ ints0 ≠ [] → n→i(# ints0) ≤ j1 - j0 ⊦ ints = [], dups(ints)

Rule argument

fma: max(ints) ∈ ints

Additional information

Used simplifier rules: max-in-01: ⊦ max(ints) ∈ ints ↔ ints ≠ []; Used formulas: max(ints) ∈ ints