Theorem 2sb6 1860

Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.)

Assertion

Ref	Expression
2sb6	⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑))

Distinct variable groups:   𝑥,𝑦,𝑧   𝑦,𝑤

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧,𝑤)

Proof of Theorem 2sb6

Step	Hyp	Ref	Expression
1		sb6 1766	. 2 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∀𝑥(𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑))
2		19.21v 1753	. . . 4 ⊢ (∀𝑦(𝑥 = 𝑧 → (𝑦 = 𝑤 → 𝜑)) ↔ (𝑥 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
3		impexp 250	. . . . 5 ⊢ (((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑) ↔ (𝑥 = 𝑧 → (𝑦 = 𝑤 → 𝜑)))
4	3	albii 1359	. . . 4 ⊢ (∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑) ↔ ∀𝑦(𝑥 = 𝑧 → (𝑦 = 𝑤 → 𝜑)))
5		sb6 1766	. . . . 5 ⊢ ([𝑤 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑤 → 𝜑))
6	5	imbi2i 215	. . . 4 ⊢ ((𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑) ↔ (𝑥 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
7	2, 4, 6	3bitr4ri 202	. . 3 ⊢ ((𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑) ↔ ∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑))
8	7	albii 1359	. 2 ⊢ (∀𝑥(𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑) ↔ ∀𝑥∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑))
9	1, 8	bitri 173	1 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑))

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  ax-i5r 1428

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

This theorem is referenced by: (None)