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Proof step 20 from proof max-exists (previous step)

ino0 + ino < n, n ∈ files0, ∀ n. n ∈ files0 → n < ino0, ∀ 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, ino ∈ files0, ∃ ino0. ino0 ≠ 0 ∧ (∀ n. n = ino ∨ n ∈ files0 → n < ino0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + m0 < n → ¬ n + n0 < m; (s) Used terms: n