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Theorem bj-dvelimdv 37940
Description: Deduction form of dvelim 2483 with disjoint variable conditions. Uncurried (imported) form of bj-dvelimdv1 37941. Typically, 𝑧 is a fresh variable used for the implicit substitution hypothesis that results in 𝜒 (namely, 𝜓 can be thought as 𝜓(𝑥, 𝑦) and 𝜒 as 𝜓(𝑥, 𝑧)). So the theorem says that if x is effectively free in 𝜓(𝑥, 𝑧), then if x and y are not the same variable, then 𝑥 is also effectively free in 𝜓(𝑥, 𝑦), in a context 𝜑.

One can weaken the implicit substitution hypothesis by adding the antecedent 𝜑 but this typically does not make the theorem much more useful. Similarly, one could use nonfreeness hypotheses instead of disjoint variable conditions but since this result is typically used when 𝑧 is a dummy variable, this would not be of much benefit. One could also remove DV (𝑥, 𝑧) since in the proof nfv 1949 can be replaced with nfal 2357 followed by nfn 1890.

Remark: nfald 2362 uses ax-11 2197; it might be possible to inline and use ax11w 2169 instead, but there is still a use via 19.12 2361 anyway. (Contributed by BJ, 20-Oct-2021.) (Proof modification is discouraged.)

Hypotheses
Ref Expression
bj-dvelimdv.nf (𝜑 → Ⅎ𝑥𝜒)
bj-dvelimdv.is (𝑧 = 𝑦 → (𝜒 ↔ 𝜓))
Assertion
Ref Expression
bj-dvelimdv ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥𝜓)
Distinct variable groups:   𝑥,𝑧   𝑦,𝑧   𝜑,𝑧   𝜓,𝑧
Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝜓(𝑥, 𝑦)   𝜒(𝑥, 𝑦, 𝑧)

Proof of Theorem bj-dvelimdv
StepHypRef Expression
1 bj-dvelimdv.is . . . 4 (𝑧 = 𝑦 → (𝜒 ↔ 𝜓))
21equsalvw 2039 . . 3 (∀𝑧(𝑧 = 𝑦 → 𝜒) ↔ 𝜓)
32bicomi 227 . 2 (𝜓 ↔ ∀𝑧(𝑧 = 𝑦 → 𝜒))
4 nfv 1949 . . . 4 Ⅎ𝑧𝜑
5 nfv 1949 . . . 4 Ⅎ𝑧 ¬ ∀𝑥 𝑥 = 𝑦
64, 5nfan 1934 . . 3 Ⅎ𝑧(𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦)
7 nfeqf2 2409 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → Ⅎ𝑥 𝑧 = 𝑦)
87adantl 487 . . . 4 ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥 𝑧 = 𝑦)
9 bj-dvelimdv.nf . . . . 5 (𝜑 → Ⅎ𝑥𝜒)
109adantr 486 . . . 4 ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥𝜒)
118, 10nfimd 1929 . . 3 ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥(𝑧 = 𝑦 → 𝜒))
126, 11nfald 2362 . 2 ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥∀𝑧(𝑧 = 𝑦 → 𝜒))
133, 12nfxfrd 1887 1 ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥𝜓)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  Ⅎwnf 1816
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 1945  ax-6 2002  ax-7 2043  ax-10 2180  ax-11 2197  ax-12 2216  ax-13 2404
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817
This theorem is used by: (None)
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