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Theorem bj-rep 37738
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 3079 . . . 4 (∀𝑦𝑥 ∃!𝑧𝜑 ↔ ∀𝑦(𝑦𝑥 → ∃!𝑧𝜑))
2 eumo 2605 . . . . . . 7 (∃!𝑧𝜑 → ∃*𝑧𝜑)
32imim2i 17 . . . . . 6 ((𝑦𝑥 → ∃!𝑧𝜑) → (𝑦𝑥 → ∃*𝑧𝜑))
4 moanimv 2646 . . . . . 6 (∃*𝑧(𝑦𝑥𝜑) ↔ (𝑦𝑥 → ∃*𝑧𝜑))
53, 4sylibr 237 . . . . 5 ((𝑦𝑥 → ∃!𝑧𝜑) → ∃*𝑧(𝑦𝑥𝜑))
65alimi 1840 . . . 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 3113 . . . . . . 7 (∃𝑦𝑥 (𝑦𝑥𝜑) ↔ ∃𝑦𝑥 𝜑)
109bibi2i 340 . . . . . 6 ((𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ (𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1110albii 1848 . . . . 5 (∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 (𝑦𝑥𝜑)) ↔ ∀𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
1211exbii 1877 . . . 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 1824 1 𝑥(∀𝑦𝑥 ∃!𝑧𝜑 → ∃𝑡𝑧(𝑧𝑡 ↔ ∃𝑦𝑥 𝜑))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wb 209  wa 400  wal 1567  wex 1808  ∃*wmo 2564  ∃!weu 2595  wral 3078  wrex 3088
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-rep 5237
This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-mo 2566  df-eu 2596  df-ral 3079  df-rex 3089
This theorem is used by: (None)
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