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Proof step 13 from proof links-dirs-union (previous step)

(m ∈ dirs0 ⊃ dirs0[m];dirs1[m]) = mkdir(m0, md, m1, m2, entries), 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 ∈ dirs1[m].entries ∧ ino = dirs1[m].entries[str]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
(m ∈ dirs0 ⊃ mkdir(m3, md0, m4, m5, entries0);dirs1[m]) = mkdir(m0, md, m1, m2, entries), 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 ∈ dirs1[m].entries ∧ ino = dirs1[m].entries[str]

Proof step 13 from proof links-dirs-union (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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