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Proof step 31 from proof max-exists (previous step)

∀ n. n ∈ dirs0 → n < ino0, ∀ 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, ino0 = 0, ino ∈ dirs0, ∃ ino0. ino0 ≠ 0 ∧ (∀ n. n = ino ∨ n ∈ dirs0 → n < ino0)

User applied rule exists right

And obtained the following new goals

Proof step 32:
∀ n. n ∈ dirs0 → n < ino0, ∀ 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, ino0 = 0, ino ∈ dirs0, ∃ ino0. ino0 ≠ 0 ∧ (∀ n. n = ino ∨ n ∈ dirs0 → n < ino0), ino0 + ino + 1 ≠ 0 ∧ (∀ n. n = ino ∨ n ∈ dirs0 → n < ino0 + ino + 1)

Rule argument

pos: right 4 terms: ino0 + ino + 1 abort: F discard: F computed: F

Additional information

Used terms: ino0 + ino + 1