Theorem wl-ax11-lem10 37061

Description: We now have prepared everything. The unwanted variable 𝑢 is just in one place left. pm2.61 191 can be used in conjunction with wl-ax11-lem9 37060 to eliminate the second antecedent. Missing is something along the lines of ax-6 1964, so we could remove the first antecedent. But the Metamath axioms cannot accomplish this. Such a rule must reside one abstraction level higher than all others: It says that a distinctor implies a distinct variable condition on its contained setvar. This is only needed if such conditions are required, as ax-11v does. The result of this study is for me, that you cannot introduce a setvar capturing this condition, and hope to eliminate it later. (Contributed by Wolf Lammen, 30-Jun-2019.)

Assertion

Ref	Expression
wl-ax11-lem10	⊢ (∀𝑦 𝑦 = 𝑢 → (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦∀𝑥𝜑 → ∀𝑥∀𝑦𝜑)))

Distinct variable group:   𝑥,𝑢

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑢)

Proof of Theorem wl-ax11-lem10

Step	Hyp	Ref	Expression
1		wl-ax11-lem8 37059	. . . . 5 ⊢ ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (∀𝑢∀𝑥[𝑢 / 𝑦]𝜑 ↔ ∀𝑦∀𝑥𝜑))
2		wl-ax11-lem6 37057	. . . . 5 ⊢ ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (∀𝑢∀𝑥[𝑢 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦𝜑))
3	1, 2	bitr3d 281	. . . 4 ⊢ ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (∀𝑦∀𝑥𝜑 ↔ ∀𝑥∀𝑦𝜑))
4	3	biimpd 228	. . 3 ⊢ ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (∀𝑦∀𝑥𝜑 → ∀𝑥∀𝑦𝜑))
5	4	ex 412	. 2 ⊢ (∀𝑢 𝑢 = 𝑦 → (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦∀𝑥𝜑 → ∀𝑥∀𝑦𝜑)))
6	5	aecoms 2423	1 ⊢ (∀𝑦 𝑦 = 𝑢 → (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦∀𝑥𝜑 → ∀𝑥∀𝑦𝜑)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395  ∀wal 1532  [wsb 2060

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-10 2130  ax-12 2167  ax-13 2367  ax-wl-11v 37051

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-nf 1779  df-sb 2061

This theorem is referenced by: (None)

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