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Proof step 6 from proof datakeys-set-add (previous step)

⊦ key0 ∈ ri ∧ key0.data? ∧ (inodekey(key0.ino) = key ∨ inodekey(key0.ino) ∈ ks) ↔ key0 ∈ datakeys(ks, ri) ∨ key0 ∈ datakeys(key, ri)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
key0 ∈ ri ∧ key0.data? ∧ inodekey(key0.ino) ∈ ks ∨ key0 ∈ ri ∧ key0.data? ∧ key.inode? ∧ key0.ino = key.ino ⊦ key0 ∈ ri ∧ key0.data? ∧ (inodekey(key0.ino) = key ∨ inodekey(key0.ino) ∈ ks)

Rule argument

Additional information

Used simplifier rules: datakeys-set: ⊦ key ∈ datakeys(ks, ri) ↔ key ∈ ri ∧ key.data? ∧ inodekey(key.ino) ∈ ks;%datakeys: ⊦ key ∈ datakeys(key0, ri) ↔ key ∈ ri ∧ key.data? ∧ key0.inode? ∧ key.ino = key0.ino;% ⊦ inodekey(m).ino = m; (s)% ⊦ inodekey(m).inode?; (s)