Theorem frege100 44907

Description: One direction of dffrege99 44906. Proposition 100 of [Frege1879] p. 72. (Contributed by RP, 7-Jul-2020.) (Proof modification is discouraged.)

Hypothesis

Ref	Expression
frege99.z	⊢ 𝑍 ∈ 𝑈

Assertion

Ref	Expression
frege100	⊢ (𝑋((t+‘𝑅) ∪ I )𝑍 → (¬ 𝑋(t+‘𝑅)𝑍 → 𝑍 = 𝑋))

Proof of Theorem frege100

Step	Hyp	Ref	Expression
1		frege99.z	. . 3 ⊢ 𝑍 ∈ 𝑈
2	1	dffrege99 44906	. 2 ⊢ ((¬ 𝑋(t+‘𝑅)𝑍 → 𝑍 = 𝑋) ↔ 𝑋((t+‘𝑅) ∪ I )𝑍)
3		frege57aid 44816	. 2 ⊢ (((¬ 𝑋(t+‘𝑅)𝑍 → 𝑍 = 𝑋) ↔ 𝑋((t+‘𝑅) ∪ I )𝑍) → (𝑋((t+‘𝑅) ∪ I )𝑍 → (¬ 𝑋(t+‘𝑅)𝑍 → 𝑍 = 𝑋)))
4	2, 3	ax-mp 5	1 ⊢ (𝑋((t+‘𝑅) ∪ I )𝑍 → (¬ 𝑋(t+‘𝑅)𝑍 → 𝑍 = 𝑋))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2145   ∪ cun 3896   class class class wbr 5102   I cid 5541  ‘cfv 6527  t+ctcl 15106

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732  ax-sep 5248  ax-pr 5390  ax-frege1 44734  ax-frege2 44735  ax-frege8 44753  ax-frege52a 44801

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ifp 1079  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-br 5103  df-opab 5167  df-id 5542  df-xp 5653  df-rel 5654

This theorem is used by:  frege101  44908  frege103  44910