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Proof step 6 from proof counttheshit (previous step)

∀ alink. ¬ alink ∈ ls, ∀ ps0, fpathtolink, ls. # ps0 < # ∅ → (∀ p, q. p ∈ ps0 ∧ q ∈ ps0 ∧ fpathtolink(p) = fpathtolink(q) → p = q) ∧ (∀ alink. alink ∈ ls → (∃ p. p ∈ ps0 ∧ fpathtolink(p) = alink)) ∧ (∀ p. p ∈ ps0 → fpathtolink(p) ∈ ls) → (# ps0 = # ls ↔ true) ⊦ 0 = # ls, ls = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ls = ls0 ↔ (∀ alink. alink ∈ ls ↔ alink ∈ ls0)
Proof step 8:
∀ alink. ¬ alink ∈ ls, ∀ ps0, fpathtolink, ls. # ps0 < # ∅ → (∀ p, q. p ∈ ps0 ∧ q ∈ ps0 ∧ fpathtolink(p) = fpathtolink(q) → p = q) ∧ (∀ alink. alink ∈ ls → (∃ p. p ∈ ps0 ∧ fpathtolink(p) = alink)) ∧ (∀ p. p ∈ ps0 → fpathtolink(p) ∈ ls) → (# ps0 = # ls ↔ true) ⊦ 0 = # ls, ∀ alink. alink ∈ ls ↔ alink ∈ ∅

Rule argument

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

Additional information

Used terms: ∅, ls