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Proof step 3 from proof links-dirs-union (previous step)

dirs0 ⊥ dirs1 ⊦ links(ino, dirs0 ∪ dirs1) = links(ino, dirs0) ∪ links(ino, dirs1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ls = ls0 ↔ (∀ alink. alink ∈ ls ↔ alink ∈ ls0)
Proof step 5:
dirs0 ⊥ dirs1 ⊦ ∀ alink. alink ∈ links(ino, dirs0 ∪ dirs1) ↔ alink ∈ links(ino, dirs0) ∪ links(ino, dirs1)

Rule argument

spec: set-basic, inst: alink-set name: Extension seq: ⊦ ls = ls0 ↔ (∀ alink. alink ∈ ls ↔ alink ∈ ls0) subst: [ls0, ls] → [links(ino, dirs0) ∪ links(ino, dirs1), links(ino, dirs0 ∪ dirs1)] flags: <all paths>

Additional information

Used terms: links(ino, dirs0) ∪ links(ino, dirs1), links(ino, dirs0 ∪ dirs1)