Theorem sbanv 1769

Description: Version of sban 1829 where 𝑥 and 𝑦 are distinct. (Contributed by Jim Kingdon, 24-Dec-2017.)

Assertion

Ref	Expression
sbanv	⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Distinct variable group:   𝑥,𝑦

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

Proof of Theorem sbanv

Step	Hyp	Ref	Expression
1		sb6 1766	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ∀𝑥(𝑥 = 𝑦 → (𝜑 ∧ 𝜓)))
2		sb6 1766	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3		sb6 1766	. . . 4 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
4	2, 3	anbi12i 433	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∧ ∀𝑥(𝑥 = 𝑦 → 𝜓)))
5		19.26 1370	. . 3 ⊢ (∀𝑥((𝑥 = 𝑦 → 𝜑) ∧ (𝑥 = 𝑦 → 𝜓)) ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∧ ∀𝑥(𝑥 = 𝑦 → 𝜓)))
6		pm4.76 536	. . . 4 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ (𝑥 = 𝑦 → 𝜓)) ↔ (𝑥 = 𝑦 → (𝜑 ∧ 𝜓)))
7	6	albii 1359	. . 3 ⊢ (∀𝑥((𝑥 = 𝑦 → 𝜑) ∧ (𝑥 = 𝑦 → 𝜓)) ↔ ∀𝑥(𝑥 = 𝑦 → (𝜑 ∧ 𝜓)))
8	4, 5, 7	3bitr2i 197	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ∀𝑥(𝑥 = 𝑦 → (𝜑 ∧ 𝜓)))
9	1, 8	bitr4i 176	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1241  [wsb 1645

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427

This theorem depends on definitions:  df-bi 110  df-sb 1646

This theorem is referenced by:  sban  1829