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Proof step 10 from proof links-inc-nsubdirs (previous step)

(m = p_ino ⊃ mkdir(m0, md, n0, n1 + 1, entries);dirs[m]) = mkdir(m1, md0, m2, m3, entries0), (m = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[m]) = mkdir(m4, md1, m5, m6, entries1), p_ino ∈ dirs, m ∈ dirs ⊦ str ∈ entries0 ∧ ino = entries0[str] ↔ str ∈ entries1 ∧ ino = entries1[str]

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 2

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)