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Proof step 12 from proof links-putnew (previous step)

(m0 = ino ⊃ mkdir(m, md, n, n0, ∅);mkdir(m4, md1, m5, m6, entries0)) = mkdir(m1, md0, m2, m3, entries), dirs[m0] = mkdir(m4, md1, m5, m6, entries0) ⊦ (m0 = ino ∨ m0 ∈ dirs) ∧ str ∈ entries ∧ ino0 = entries[str] ↔ m0 ∈ dirs ∧ str ∈ entries0 ∧ ino0 = entries0[str], ino = 0, ino ∈ dirs

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)% ⊦ ¬ str ∈ ∅; (s)