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Proof step 25 from proof new-inode-exists (previous step)

ino + m < ino0, inodekey(ino0) ∈ ks0, ∀ ino0. inodekey(ino0) ∈ ks0 → ino0 < ino, ∀ ks1. # ks1 < #(ks0 ++ inodekey(m)) → true → (∃ ino. ∀ ino0. inodekey(ino0) ∈ ks1 → ino0 < ino) ⊦ # ks0 = 0, ks0 = ∅, inodekey(m) ∈ ks0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: ino0

Additional information

Used simplifier rules: 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 < n → n ≠ 0; (s)% ⊦ 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: ino0