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Proof step 19 from proof max-exists (previous step)

∀ 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), ino0 + ino + 1 ≠ 0 ∧ (∀ n. n = ino ∨ n ∈ files0 → n < ino0 + ino + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
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)

Rule argument

Additional information

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