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Proof step 1662 from proof inv-renam (
previous step
)
p = p1, p0 = p2, dom of < ino, afs-cons(dirs, files), of-cons(of, dirs, files) ⊦ ∃ old_dent, new_dent, isdir, inode, old_inode, new_inode, p_inode, old_dent_inode, new_dent_inode. true
Heuristic simplifier applied rule simplifier
And closed the goal!
Rule argument
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