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

dirs = ∅, ∀ dirs0, files0. # dirs0 < # dirs → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs0 ∨ n ∈ files0 → n < ino)), ∀ files0, dirs. # files0 < # files → (∀ dirs0, files. # dirs0 < # dirs → true → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs0 ∨ n ∈ files → n < ino))) → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs ∨ n ∈ files0 → n < ino)) ⊦ ∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs ∨ n ∈ files → n < ino)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ files0, dirs. # files0 < # files → (∀ dirs0, files. # dirs0 < # dirs → true → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs0 ∨ n ∈ files → n < ino))) → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs ∨ n ∈ files0 → n < ino)) ⊦ ∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ files → n < ino)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ino ∈ ∅; (s)% ⊦ ¬ m < 0; (s)% ⊦ # ∅ = 0; (s)