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Proof step 2 from proof abs-keys-orphans-add (previous step)

ro.inodes? ⊦ abs-keys(ro) = abs-keys(orphans(fs, ri)) ++ n ↔ ro = orphans(fs, ri) ++ inodekey(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
abs-keys(ro) = abs-keys(orphans(fs, ri)) ++ n, ro.inodes? ⊦ ro = orphans(fs, ri) ++ inodekey(n)

Rule argument

Additional information

Used simplifier rules: abs-keys-add: ⊦ abs-keys(ks ++ inodekey(ino)) = abs-keys(ks) ++ ino;