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Proof step 28 from proof max-exists (previous step)

∀ dirs0, files0. # dirs0 < # dirs → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs0 ∨ n ∈ files0 → n < ino)), ∃ dirs0, ad. dirs = (dirs0 ++ ino)[ino, ad] ∧ ¬ ino ∈ dirs0, ∀ files0, dirs. # files0 < # ∅ → (∀ 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 = 0, ∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs → n < ino)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
∀ dirs, files. ¬ # dirs0 < # dirs → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs ∨ n ∈ files → n < ino)), ∀ files0, dirs. # files0 < # ∅ → (∀ 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 = 0, ino ∈ dirs0, ∃ ino0. ino0 ≠ 0 ∧ (∀ n. n = ino ∨ n ∈ dirs0 → n < ino0)

Rule argument

Additional information

Used simplifier rules: ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%¬ ino ∈ dirs, ino ≠ 0 ⊦ #(dirs ++ ino) = # dirs + 1; (s)%ino ∈ dirs ⊦ # dirs[ino, ad] = # dirs; (s)%ino ≠ 0, ¬ ino ∈ dirs ⊦ ino0 ∈ dirs ++ ino ↔ ino0 = ino ∨ ino0 ∈ dirs; (s)