Theorem 2sb6 2089

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 2088	. 2 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∀𝑥(𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑))
2		19.21v 1942	. . . 4 ⊢ (∀𝑦(𝑥 = 𝑧 → (𝑦 = 𝑤 → 𝜑)) ↔ (𝑥 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
3		impexp 451	. . . . 5 ⊢ (((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑) ↔ (𝑥 = 𝑧 → (𝑦 = 𝑤 → 𝜑)))
4	3	albii 1822	. . . 4 ⊢ (∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑) ↔ ∀𝑦(𝑥 = 𝑧 → (𝑦 = 𝑤 → 𝜑)))
5		sb6 2088	. . . . 5 ⊢ ([𝑤 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑤 → 𝜑))
6	5	imbi2i 336	. . . 4 ⊢ ((𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑) ↔ (𝑥 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
7	2, 4, 6	3bitr4ri 304	. . 3 ⊢ ((𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑) ↔ ∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑))
8	7	albii 1822	. 2 ⊢ (∀𝑥(𝑥 = 𝑧 → [𝑤 / 𝑦]𝜑) ↔ ∀𝑥∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑))
9	1, 8	bitri 274	1 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) → 𝜑))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1537  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011

This theorem depends on definitions:  df-bi 206  df-an 397  df-ex 1783  df-sb 2068

This theorem is referenced by:  sbcom2  2161  2exsb  2358  2eu6  2658