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Proof step 13 from proof counttheshit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
p ∈ ps, ∀ p, q. p ∈ ps ∧ q ∈ ps ∧ fpathtolink(p) = fpathtolink(q) → p = q, ∀ alink. alink ∈ ls → (∃ p. p ∈ ps ∧ fpathtolink(p) = alink), ∀ p. p ∈ ps → fpathtolink(p) ∈ ls, ∀ ps0, fpathtolink, ls. # ps0 < # ps → (∀ 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) ⊦ # ps = 0, # ps = # ls

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ p ∈ ∅; (s)