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Proof step 7 from proof subdirs-add-entry (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
ino ∈ dirs, str ∈ entries, entries[str] ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])]) = subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) -- str, ino0 ∈ dirs
Proof step 12:
ino ∈ dirs, ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])]) = subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]), ino0 ∈ dirs

Rule argument

pos: right 1

Proof step 7 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