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Theorem bj-rep 37378
Description: Version of the axiom of replacement requiring the functional relation in the axiom to be a (total) function from ax-rep 5212 (in the form of axrep6 5221). (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 3053 . . . 4 (∀𝑦𝑥 ∃!𝑧𝜑 ↔ ∀𝑦(𝑦𝑥 → ∃!𝑧𝜑))
2 eumo 2579 . . . . . . 7 (∃!𝑧𝜑 → ∃*𝑧𝜑)
32imim2i 16 . . . . . 6 ((𝑦𝑥 → ∃!𝑧𝜑) → (𝑦𝑥 → ∃*𝑧𝜑))
4 moanimv 2620 . . . . . 6 (∃*𝑧(𝑦𝑥𝜑) ↔ (𝑦𝑥 → ∃*𝑧𝜑))
53, 4sylibr 234 . . . . 5 ((𝑦𝑥 → ∃!𝑧𝜑) → ∃*𝑧(𝑦𝑥𝜑))
65alimi 1813 . . . 4 (∀𝑦(𝑦𝑥 → ∃!𝑧𝜑) → ∀𝑦∃*𝑧(𝑦𝑥𝜑))
71, 6sylbi 217 . . 3 (∀𝑦𝑥 ∃!𝑧𝜑 → ∀𝑦∃*𝑧(𝑦𝑥𝜑))
8 axrep6 5221 . . . 4 (∀𝑦∃*𝑧(𝑦𝑥𝜑) → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)))
9 rexanid 3087 . . . . . . 7 (∃𝑦𝑥 (𝑦𝑥𝜑) ↔ ∃𝑦𝑥 𝜑)
109bibi2i 337 . . . . . 6 ((𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ (𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1110albii 1821 . . . . 5 (∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ ∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1211exbii 1850 . . . 4 (∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
138, 12sylib 218 . . 3 (∀𝑦∃*𝑧(𝑦𝑥𝜑) → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
147, 13syl 17 . 2 (∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1514ax-gen 1797 1 𝑥(∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wal 1540  wex 1781  ∃*wmo 2538  ∃!weu 2569  wral 3052  wrex 3062
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-rep 5212
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-mo 2540  df-eu 2570  df-ral 3053  df-rex 3063
This theorem is referenced by: (None)
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