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Proof step 53 from proof max-exists (previous step)

∀ n. n ∈ dirs0 ∨ n ∈ files0 → n < ino1, ∀ n. n ∈ dirs0 ∨ n ∈ files0 → n < ino2, ∀ files1, dirs. # files1 < # (files0 ++ ino)[ino, af] → (∀ dirs0, files. # dirs0 < # dirs → true → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs0 ∨ n ∈ files → n < ino))) → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs ∨ n ∈ files1 → n < ino)) ⊦ ino = 0, ino0 = 0, ino1 = 0, ino2 = 0, ino0 ∈ dirs0, ino ∈ files0, ∃ ino1. ino1 ≠ 0 ∧ (∀ n. n = ino0 ∨ n ∈ dirs0 ∨ n = ino ∨ n ∈ files0 → n < ino1)

User applied rule exists right

And obtained the following new goals

Proof step 54:
∀ n. n ∈ dirs0 ∨ n ∈ files0 → n < ino2, ∀ n. n ∈ dirs0 ∨ n ∈ files0 → n < ino1, ∀ files1, dirs. # files1 < # (files0 ++ ino)[ino, af] → (∀ dirs0, files. # dirs0 < # dirs → true → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs0 ∨ n ∈ files → n < ino))) → (∃ ino. ino ≠ 0 ∧ (∀ n. n ∈ dirs ∨ n ∈ files1 → n < ino)) ⊦ ino = 0, ino0 = 0, ino1 = 0, ino2 = 0, ino ∈ files0, ino0 ∈ dirs0, ∃ ino1. ino1 ≠ 0 ∧ (∀ n. n = ino0 ∨ n ∈ dirs0 ∨ n = ino ∨ n ∈ files0 → n < ino1), ino0 + ino + ino1 + 1 ≠ 0 ∧ (∀ n. n = ino0 ∨ n ∈ dirs0 ∨ n = ino ∨ n ∈ files0 → n < ino0 + ino + ino1 + 1)

Rule argument

pos: right 7 terms: ino0 + ino + ino1 + 1 abort: F discard: F computed: F

Additional information

Used terms: ino0 + ino + ino1 + 1