Theorem sbel2x 1347

Description: Elimination of double substitution.

Assertion

Ref	Expression
sbel2x	⊢ (φ ↔ ∃x∃y((x = z ⋀ y = w) ⋀ [y / w][x / z]φ))

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

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 1346	. . . . 5 ⊢ ([x / z]φ ↔ ∃y(y = w ⋀ [y / w][x / z]φ))
2	1	anbi2i 482	. . . 4 ⊢ ((x = z ⋀ [x / z]φ) ↔ (x = z ⋀ ∃y(y = w ⋀ [y / w][x / z]φ)))
3	2	exbii 1053	. . 3 ⊢ (∃x(x = z ⋀ [x / z]φ) ↔ ∃x(x = z ⋀ ∃y(y = w ⋀ [y / w][x / z]φ)))
4		sbelx 1346	. . 3 ⊢ (φ ↔ ∃x(x = z ⋀ [x / z]φ))
5		exdistr 1311	. . 3 ⊢ (∃x∃y(x = z ⋀ (y = w ⋀ [y / w][x / z]φ)) ↔ ∃x(x = z ⋀ ∃y(y = w ⋀ [y / w][x / z]φ)))
6	3, 4, 5	3bitr4 183	. 2 ⊢ (φ ↔ ∃x∃y(x = z ⋀ (y = w ⋀ [y / w][x / z]φ)))
7		anass 441	. . 3 ⊢ (((x = z ⋀ y = w) ⋀ [y / w][x / z]φ) ↔ (x = z ⋀ (y = w ⋀ [y / w][x / z]φ)))
8	7	2exbii 1054	. 2 ⊢ (∃x∃y((x = z ⋀ y = w) ⋀ [y / w][x / z]φ) ↔ ∃x∃y(x = z ⋀ (y = w ⋀ [y / w][x / z]φ)))
9	6, 8	bitr4 176	1 ⊢ (φ ↔ ∃x∃y((x = z ⋀ y = w) ⋀ [y / w][x / z]φ))

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

This theorem is referenced by:  opabid 2816

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  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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