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Proof step 24 from proof counttheshit (previous step)

∀ p0, q. (p0 = p ∨ p0 ∈ ps0) ∧ (q = p ∨ q ∈ ps0) ∧ fpathtolink(p0) = fpathtolink(q) → p0 = q, ∀ p0. p0 ∈ ps0 → fpathtolink(p0) = fpathtolink(p) ∨ fpathtolink(p0) ∈ ls0, ∀ alink. alink = fpathtolink(p) ∨ alink ∈ ls0 → (∃ p0. (p0 = p ∨ p0 ∈ ps0) ∧ fpathtolink(p0) = alink), ∀ 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) ⊦ # ps0 = # ls0, p ∈ ps0, fpathtolink(p) ∈ ls0, (p0 ∈ ps0 ∧ q ∈ ps0 ∧ fpathtolink(p0) = fpathtolink(q) → p0 = q) ∧ (alink ∈ ls0 → (∃ p. p ∈ ps0 ∧ fpathtolink(p) = alink)) ∧ (p1 ∈ ps0 → fpathtolink(p1) ∈ ls0)

User applied rule case distinction

And obtained the following new goals

Proof step 25:
fpathtolink(p0) = fpathtolink(q), p0 ∈ ps0, q ∈ ps0, ∀ alink. alink = fpathtolink(p) ∨ alink ∈ ls0 → (∃ p0. (p0 = p ∨ p0 ∈ ps0) ∧ fpathtolink(p0) = alink), ∀ p0. p0 ∈ ps0 → fpathtolink(p0) = fpathtolink(p) ∨ fpathtolink(p0) ∈ ls0, ∀ 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) ⊦ # ps0 = # ls0, p0 = q, fpathtolink(p) ∈ ls0, p ∈ ps0
Proof step 27:
alink ∈ ls0, ∀ alink. alink = fpathtolink(p) ∨ alink ∈ ls0 → (∃ p0. (p0 = p ∨ p0 ∈ ps0) ∧ fpathtolink(p0) = alink), ∀ p0. p0 ∈ ps0 → fpathtolink(p0) = fpathtolink(p) ∨ fpathtolink(p0) ∈ ls0, ∀ 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) ⊦ # ps0 = # ls0, fpathtolink(p) ∈ ls0, p ∈ ps0, ∃ p. p ∈ ps0 ∧ fpathtolink(p) = alink
Proof step 31:
p1 ∈ ps0, ∀ alink. alink = fpathtolink(p) ∨ alink ∈ ls0 → (∃ p0. (p0 = p ∨ p0 ∈ ps0) ∧ fpathtolink(p0) = alink), ∀ p0. p0 ∈ ps0 → fpathtolink(p0) = fpathtolink(p) ∨ fpathtolink(p0) ∈ ls0, ∀ 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) ⊦ # ps0 = # ls0, fpathtolink(p) ∈ ls0, p ∈ ps0, fpathtolink(p1) ∈ ls0

Rule argument

pos: right 4

Proof step 24 from proof counttheshit (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information