Theorem sbel2x 995

Description: Elimination of double substitution.

Assertion

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

Distinct variable group(s):   x,y,z,w   φ,x,y

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 994	. . . . 5 ⊢ ([x / z]φ ↔ ∃y(y = w ∧ [y / w][x / z]φ))
2	1	anbi2i 367	. . . 4 ⊢ ((x = z ∧ [x / z]φ) ↔ (x = z ∧ ∃y(y = w ∧ [y / w][x / z]φ)))
3	2	biex 733	. . 3 ⊢ (∃x(x = z ∧ [x / z]φ) ↔ ∃x(x = z ∧ ∃y(y = w ∧ [y / w][x / z]φ)))
4		sbelx 994	. . 3 ⊢ (φ ↔ ∃x(x = z ∧ [x / z]φ))
5		exdistr 967	. . 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 158	. 2 ⊢ (φ ↔ ∃x∃y(x = z ∧ (y = w ∧ [y / w][x / z]φ)))
7		anass 336	. . 3 ⊢ (((x = z ∧ y = w) ∧ [y / w][x / z]φ) ↔ (x = z ∧ (y = w ∧ [y / w][x / z]φ)))
8	7	bi2ex 734	. 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 154	1 ⊢ (φ ↔ ∃x∃y((x = z ∧ y = w) ∧ [y / w][x / z]φ))

Syntax hints:   ↔ wb 127   ∧ wa 196  ∃wex 678   = weq 797  [wsb 852

This theorem is referenced by:  opabid 2099

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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