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Proof step 22 from proof links-del-entry (previous step)

(m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries -- str);dirs[m0]) = mkdir(m1, md0, m2, m3, entries0), (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries);dirs[m0]) = mkdir(m4, md1, m5, m6, entries1), ino0 ∈ dirs, str ∈ entries, m0 ∈ dirs ⊦ str0 ∈ entries0 ∧ ino = entries0[str0] ↔ str0 ∈ entries1 ∧ ino = entries1[str0], entries[str] = ino

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: m0 = ino0

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 ∈ entries -- str0 ↔ str ≠ str0 ∧ str ∈ entries; (s)% ⊦ str ≠ str0 → (entries -- str0)[str] = entries[str]; (s)%⊦ entries[str] ≠ entries[str0] → (str ∈ entries -- str0 ↔ str ∈ entries); (s)%str ≠ str0 ⊦ (entries -- str0)[str] = entries[str]; (s) Used formulas: m0 = ino0