Theorem bj-ssblem2 37681

Description: An instance of ax-11 2196 proved without it. The converse may not be provable without ax-11 2196 (since using alcomimw 2077 would require a DV on 𝜑, 𝑥, which defeats the purpose). (Contributed by BJ, 22-Dec-2020.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-ssblem2	⊢ (∀𝑥∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦 → 𝜑)) → ∀𝑦∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦 → 𝜑)))

Distinct variable groups:   𝑥,𝑦,𝑡   𝜑,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑡)

Proof of Theorem bj-ssblem2

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ1 2059	. . 3 ⊢ (𝑦 = 𝑧 → (𝑦 = 𝑡 ↔ 𝑧 = 𝑡))
2		equequ2 2060	. . . 4 ⊢ (𝑦 = 𝑧 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑧))
3	2	imbi1d 344	. . 3 ⊢ (𝑦 = 𝑧 → ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑧 → 𝜑)))
4	1, 3	imbi12d 347	. 2 ⊢ (𝑦 = 𝑧 → ((𝑦 = 𝑡 → (𝑥 = 𝑦 → 𝜑)) ↔ (𝑧 = 𝑡 → (𝑥 = 𝑧 → 𝜑))))
5	4	alcomimw 2077	1 ⊢ (∀𝑥∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦 → 𝜑)) → ∀𝑦∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦 → 𝜑)))

This proof depends on syntax axioms:   → wi 4  ∀wal 1567

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812

This theorem is used by: (None)