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Theorem bj-rep 37654
Description: Version of the axiom of replacement requiring the functional relation in the axiom to be a (total) function from ax-rep 5237 (in the form of axrep6 5246). (Contributed by BJ, 14-Mar-2026.) The proof proves the statement without the DV condition on 𝑥, 𝜑, but the DV condition is added to this statement to show that this weaker version is sufficient. (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
bj-rep 𝑥(∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
Distinct variable groups:   𝑥,𝑦,𝑧,𝑡   𝜑,𝑥,𝑡
Allowed substitution hints:   𝜑(𝑦,𝑧)

Proof of Theorem bj-rep
StepHypRef Expression
1 df-ral 3078 . . . 4 (∀𝑦𝑥 ∃!𝑧𝜑 ↔ ∀𝑦(𝑦𝑥 → ∃!𝑧𝜑))
2 eumo 2604 . . . . . . 7 (∃!𝑧𝜑 → ∃*𝑧𝜑)
32imim2i 17 . . . . . 6 ((𝑦𝑥 → ∃!𝑧𝜑) → (𝑦𝑥 → ∃*𝑧𝜑))
4 moanimv 2645 . . . . . 6 (∃*𝑧(𝑦𝑥𝜑) ↔ (𝑦𝑥 → ∃*𝑧𝜑))
53, 4sylibr 237 . . . . 5 ((𝑦𝑥 → ∃!𝑧𝜑) → ∃*𝑧(𝑦𝑥𝜑))
65alimi 1839 . . . 4 (∀𝑦(𝑦𝑥 → ∃!𝑧𝜑) → ∀𝑦∃*𝑧(𝑦𝑥𝜑))
71, 6sylbi 220 . . 3 (∀𝑦𝑥 ∃!𝑧𝜑 → ∀𝑦∃*𝑧(𝑦𝑥𝜑))
8 axrep6 5246 . . . 4 (∀𝑦∃*𝑧(𝑦𝑥𝜑) → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)))
9 rexanid 3112 . . . . . . 7 (∃𝑦𝑥 (𝑦𝑥𝜑) ↔ ∃𝑦𝑥 𝜑)
109bibi2i 340 . . . . . 6 ((𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ (𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1110albii 1847 . . . . 5 (∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ ∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1211exbii 1876 . . . 4 (∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
138, 12sylib 221 . . 3 (∀𝑦∃*𝑧(𝑦𝑥𝜑) → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
147, 13syl 18 . 2 (∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1514ax-gen 1823 1 𝑥(∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 400  wal 1566  wex 1807  ∃*wmo 2563  ∃!weu 2594  wral 3077  wrex 3087
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-rep 5237
This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1808  df-mo 2565  df-eu 2595  df-ral 3078  df-rex 3088
This theorem is referenced by: (None)
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