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Proof step 2 from proof subdirs-add-entry (previous step)

ino ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])]) = (ino0 ∈ dirs ⊃ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) ++ str ;(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)]) ) )

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 3:
ino ∈ dirs, ino0 ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])]) = subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) ++ str
Proof step 7:
ino ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])]) = (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)]) ), ino0 ∈ dirs

Rule argument

pos: right 1

Proof step 2 from proof subdirs-add-entry (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information