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Proof step 3 from proof subdirs-del-entry (previous step)

ino ∈ dirs, str ∈ entries ⊦ 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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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)]))

Proof step 3 from proof subdirs-del-entry (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information