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Proof step 8 from proof subdirs-inc-size (previous step)

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

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)