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Proof step 19 from proof new-inode-exists (previous step)

∀ ino0. inodekey(ino0) ∈ ks0 → ino0 < ino, ∀ ks1. # ks1 < #(ks0 ++ key) → true → (∃ ino. ∀ ino0. inodekey(ino0) ∈ ks1 → ino0 < ino) ⊦ key ∈ ks0, ∃ ino. ∀ ino0. inodekey(ino0) = key ∨ inodekey(ino0) ∈ ks0 → ino0 < ino

User applied rule cut formula

And obtained the following new goals

Proof step 20:
key.inode?, ∀ ino0. inodekey(ino0) ∈ ks0 → ino0 < ino, ∀ ks1. # ks1 < #(ks0 ++ key) → true → (∃ ino. ∀ ino0. inodekey(ino0) ∈ ks1 → ino0 < ino) ⊦ key ∈ ks0, ∃ ino. ∀ ino0. inodekey(ino0) = key ∨ inodekey(ino0) ∈ ks0 → ino0 < ino
Proof step 26:
∀ ino0. inodekey(ino0) ∈ ks0 → ino0 < ino, ∀ ks1. # ks1 < #(ks0 ++ key) → true → (∃ ino. ∀ ino0. inodekey(ino0) ∈ ks1 → ino0 < ino) ⊦ key.inode?, key ∈ ks0, ∃ ino. ∀ ino0. inodekey(ino0) = key ∨ inodekey(ino0) ∈ ks0 → ino0 < ino

Rule argument

fma: key.inode?

Additional information

Used formulas: key.inode?