Theorem sb5 988

Description: Equivalence for substitution. Similar to Theorem 6.1 of [Quine] p. 40.

Assertion

Ref	Expression
sb5	⊢ ([y / x]φ ↔ ∃x(x = y ∧ φ))

Distinct variable group(s):   x,y

Proof of Theorem sb5

Step	Hyp	Ref	Expression
1		hbs1 986	. . 3 ⊢ ([y / x]φ → ∀x[y / x]φ)
2		sbequ12 865	. . 3 ⊢ (x = y → (φ ↔ [y / x]φ))
3	1, 2	eqsex 834	. 2 ⊢ (∃x(x = y ∧ φ) ↔ [y / x]φ)
4	3	bicomi 150	1 ⊢ ([y / x]φ ↔ ∃x(x = y ∧ φ))

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

This theorem is referenced by:  sb7 991  sbelx 994

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

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

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