Theorem 2sb5 1337

Description: Equivalence for double substitution.

Assertion

Ref	Expression
2sb5	⊢ ([z / x][w / y]φ ↔ ∃x∃y((x = z ⋀ y = w) ⋀ φ))

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

Proof of Theorem 2sb5

Step	Hyp	Ref	Expression
1		sb5 1270	. 2 ⊢ ([z / x][w / y]φ ↔ ∃x(x = z ⋀ [w / y]φ))
2		19.42v 1310	. . . 4 ⊢ (∃y(x = z ⋀ (y = w ⋀ φ)) ↔ (x = z ⋀ ∃y(y = w ⋀ φ)))
3		anass 441	. . . . 5 ⊢ (((x = z ⋀ y = w) ⋀ φ) ↔ (x = z ⋀ (y = w ⋀ φ)))
4	3	exbii 1053	. . . 4 ⊢ (∃y((x = z ⋀ y = w) ⋀ φ) ↔ ∃y(x = z ⋀ (y = w ⋀ φ)))
5		sb5 1270	. . . . 5 ⊢ ([w / y]φ ↔ ∃y(y = w ⋀ φ))
6	5	anbi2i 482	. . . 4 ⊢ ((x = z ⋀ [w / y]φ) ↔ (x = z ⋀ ∃y(y = w ⋀ φ)))
7	2, 4, 6	3bitr4r 184	. . 3 ⊢ ((x = z ⋀ [w / y]φ) ↔ ∃y((x = z ⋀ y = w) ⋀ φ))
8	7	exbii 1053	. 2 ⊢ (∃x(x = z ⋀ [w / y]φ) ↔ ∃x∃y((x = z ⋀ y = w) ⋀ φ))
9	1, 8	bitr 173	1 ⊢ ([z / x][w / y]φ ↔ ∃x∃y((x = z ⋀ y = w) ⋀ φ))

Syntax hints:   ↔ wb 146   ⋀ wa 223   = wceq 958  ∃wex 982  [wsbc 1172

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 965  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-16 1212  ax-11o 1220

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174

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