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Proof step 5 from proof inodes-put (previous step)

ks.inodes? ⊦ (∀ key0. key0 ∈ ks ++ key → key0.inode?) ↔ key.inode?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
ks.inodes?, ∀ key0. key0 = key ∨ key0 ∈ ks → key0.inode? ⊦ key.inode?

Rule argument

Additional information

Used simplifier rules: inodes-in: ⊦ key ∈ ks ∧ ks.inodes? → key.inode?;% ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)