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Proof step 16 from proof links-dirs-union (previous step)

(m ∈ dirs0 ⊃ mkdir(m3, md0, m4, m5, entries0);mkdir(m6, md1, m7, m8, entries1)) = mkdir(m0, md, m1, m2, entries), dirs1[m] = mkdir(m6, md1, m7, m8, entries1), dirs0[m] = mkdir(m3, md0, m4, m5, entries0), dirs0 ⊥ dirs1 ⊦ (m ∈ dirs0 ∨ m ∈ dirs1) ∧ str ∈ entries ∧ ino = entries[str] ↔ m ∈ dirs0 ∧ str ∈ entries0 ∧ ino = entries0[str] ∨ m ∈ dirs1 ∧ str ∈ entries1 ∧ ino = entries1[str]

User 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)% ⊦ ino ∈ dirs ∧ dirs ⊥ dirs0 → (ino ∈ dirs0 ↔ false); (s)