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Proof step 19 from proof counttheshit (previous step)

fpathtolink(p) ∈ ls, ∀ p0. p0 ∈ ps0 → fpathtolink(p0) ∈ ls, ∀ alink. alink ∈ ls → (∃ p0. (p0 = p ∨ p0 ∈ ps0) ∧ fpathtolink(p0) = alink), ∀ p0, q. (p0 = p ∨ p0 ∈ ps0) ∧ (q = p ∨ q ∈ ps0) ∧ fpathtolink(p0) = fpathtolink(q) → p0 = q, ∀ ps1, fpathtolink, ls. # ps1 < #(ps0 ++ p) → (∀ p, q. p ∈ ps1 ∧ q ∈ ps1 ∧ fpathtolink(p) = fpathtolink(q) → p = q) ∧ (∀ alink. alink ∈ ls → (∃ p. p ∈ ps1 ∧ fpathtolink(p) = alink)) ∧ (∀ p. p ∈ ps1 → fpathtolink(p) ∈ ls) → (# ps1 = # ls ↔ true) ⊦ # ls = 0, # ps0 + 1 = # ls, ls = ∅, p ∈ ps0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ alink ∈ ls ↔ (∃ ls0. ls = ls0 ++ alink ∧ ¬ alink ∈ ls0)
Proof step 21:
∀ p0, q. (p0 = p ∨ p0 ∈ ps0) ∧ (q = p ∨ q ∈ ps0) ∧ fpathtolink(p0) = fpathtolink(q) → p0 = q, ∀ alink. alink ∈ ls → (∃ p0. (p0 = p ∨ p0 ∈ ps0) ∧ fpathtolink(p0) = alink), ∀ p0. p0 ∈ ps0 → fpathtolink(p0) ∈ ls, ∃ ls0. ls = ls0 ++ fpathtolink(p) ∧ ¬ fpathtolink(p) ∈ ls0, ∀ ps1, fpathtolink, ls. # ps1 < #(ps0 ++ p) → (∀ p, q. p ∈ ps1 ∧ q ∈ ps1 ∧ fpathtolink(p) = fpathtolink(q) → p = q) ∧ (∀ alink. alink ∈ ls → (∃ p. p ∈ ps1 ∧ fpathtolink(p) = alink)) ∧ (∀ p. p ∈ ps1 → fpathtolink(p) ∈ ls) → (# ps1 = # ls ↔ true) ⊦ # ls = 0, # ps0 + 1 = # ls, ls = ∅, p ∈ ps0

Rule argument

spec: set-basic, inst: alink-set name: indef seq: ⊦ alink ∈ ls ↔ (∃ ls0. ls = ls0 ++ alink ∧ ¬ alink ∈ ls0) subst: [ls, alink] → [ls, fpathtolink(p)] flags: <all paths>

Additional information

Used terms: ls, fpathtolink(p)