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Proof step 4 from proof subdirs-del-entry (previous step)

str ∈ entries, ino ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries -- str)]) = (entries[str] ∈ dirs ⊃ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) -- str;subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
ino ∈ dirs, str ∈ entries, entries[str] ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries -- str)]) = subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) -- str
Proof step 7:
ino ∈ dirs, str ∈ entries ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries -- str)]) = subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]), entries[str] ∈ dirs

Rule argument

pos: right 1

Proof step 4 from proof subdirs-del-entry (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information