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Theorem bj-rep 37496
Description: Version of the axiom of replacement requiring the functional relation in the axiom to be a (total) function from ax-rep 5217 (in the form of axrep6 5226). (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 3067 . . . 4 (∀𝑦𝑥 ∃!𝑧𝜑 ↔ ∀𝑦(𝑦𝑥 → ∃!𝑧𝜑))
2 eumo 2595 . . . . . . 7 (∃!𝑧𝜑 → ∃*𝑧𝜑)
32imim2i 16 . . . . . 6 ((𝑦𝑥 → ∃!𝑧𝜑) → (𝑦𝑥 → ∃*𝑧𝜑))
4 moanimv 2636 . . . . . 6 (∃*𝑧(𝑦𝑥𝜑) ↔ (𝑦𝑥 → ∃*𝑧𝜑))
53, 4sylibr 236 . . . . 5 ((𝑦𝑥 → ∃!𝑧𝜑) → ∃*𝑧(𝑦𝑥𝜑))
65alimi 1821 . . . 4 (∀𝑦(𝑦𝑥 → ∃!𝑧𝜑) → ∀𝑦∃*𝑧(𝑦𝑥𝜑))
71, 6sylbi 219 . . 3 (∀𝑦𝑥 ∃!𝑧𝜑 → ∀𝑦∃*𝑧(𝑦𝑥𝜑))
8 axrep6 5226 . . . 4 (∀𝑦∃*𝑧(𝑦𝑥𝜑) → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)))
9 rexanid 3101 . . . . . . 7 (∃𝑦𝑥 (𝑦𝑥𝜑) ↔ ∃𝑦𝑥 𝜑)
109bibi2i 339 . . . . . 6 ((𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ (𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1110albii 1829 . . . . 5 (∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ ∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1211exbii 1858 . . . 4 (∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
138, 12sylib 220 . . 3 (∀𝑦∃*𝑧(𝑦𝑥𝜑) → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
147, 13syl 17 . 2 (∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1514ax-gen 1805 1 𝑥(∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 398  wal 1548  wex 1789  ∃*wmo 2554  ∃!weu 2585  wral 3066  wrex 3076
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1805  ax-4 1819  ax-5 1920  ax-6 1977  ax-7 2018  ax-rep 5217
This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1790  df-mo 2556  df-eu 2586  df-ral 3067  df-rex 3077
This theorem is referenced by: (None)
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