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Proof step 9 from proof links-dirs-union (previous step)

(m ∈ dirs0 ⊃ dirs0[m];dirs1[m]) = mkdir(m0, md, m1, m2, entries), dirs0 ⊥ dirs1 ⊦ (m ∈ dirs0 ∨ m ∈ dirs1) ∧ str ∈ entries ∧ ino = entries[str] ↔ mklink(m, str) ∈ links(ino, dirs0) ∪ links(ino, dirs1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ alink ∈ ls ∪ ls0 ↔ alink ∈ ls ∨ alink ∈ ls0
Proof step 11:
(m ∈ dirs0 ⊃ dirs0[m];dirs1[m]) = mkdir(m0, md, m1, m2, entries), dirs0 ⊥ dirs1 ⊦ (m ∈ dirs0 ∨ m ∈ dirs1) ∧ str ∈ entries ∧ ino = entries[str] ↔ mklink(m, str) ∈ links(ino, dirs0) ∨ mklink(m, str) ∈ links(ino, dirs1)

Rule argument

spec: set-union, inst: alink-set name: Union seq: ⊦ alink ∈ ls ∪ ls0 ↔ alink ∈ ls ∨ alink ∈ ls0 subst: [ls0, ls, alink] → [links(ino, dirs1), links(ino, dirs0), mklink(m, str)] flags: <all paths>

Additional information

Used terms: links(ino, dirs1), links(ino, dirs0), mklink(m, str)