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Proof step 5 from proof subdirs-del-mod-size (previous step)

dirs[p_ino] = mkdir(m, md, n0, n1, entries), p_ino ∈ dirs ⊦ str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) -- str ;subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) ) ↔ str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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