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Proof step 11 from proof link-mod-dir-meta (previous step)

(m0 = m ⊃ dirs[m].meta:= md;mkdir(m4, md1, m5, m6, entries0)) = mkdir(m1, md0, m2, m3, entries), dirs[m0] = mkdir(m4, md1, m5, m6, entries0), m0 ∈ dirs, m ∈ dirs ⊦ str ∈ entries ∧ ino = entries[str] ↔ str ∈ entries0 ∧ ino = entries0[str]

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)%⊦ mkdir(m, md, m0, m1, entries).meta:= md0 = mkdir(m, md0, m0, m1, entries); (s)