Theorem frege100 44805

Description: One direction of dffrege99 44804. 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 44804	. 2 ⊢ ((¬ 𝑋(t+‘𝑅)𝑍 → 𝑍 = 𝑋) ↔ 𝑋((t+‘𝑅) ∪ I )𝑍)
3		frege57aid 44714	. 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 3900   class class class wbr 5107   I cid 5553  ‘cfv 6537  t+ctcl 15060

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 2734  ax-sep 5255  ax-pr 5402  ax-frege1 44632  ax-frege2 44633  ax-frege8 44651  ax-frege52a 44699

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 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-br 5108  df-opab 5172  df-id 5554  df-xp 5665  df-rel 5666

This theorem is used by:  frege101  44806  frege103  44808