Theorem sb7f 2120

Description: This version of dfsb7 2119 does not require that φ and z be disjoint. This permits it to be used as a definition for substitution in a formalization that omits the logically redundant axiom ax-17 1616, i.e., that does not have the concept of a variable not occurring in a formula. (Contributed by NM, 26-Jul-2006.) (Revised by Mario Carneiro, 6-Oct-2016.)

Hypothesis

Ref	Expression
sb7f.1	⊢ Ⅎzφ

Assertion

Ref	Expression
sb7f	⊢ ([y / x]φ ↔ ∃z(z = y ∧ ∃x(x = z ∧ φ)))

Distinct variable groups:   x,z   y,z

Allowed substitution hints:   φ(x, y, z)

Proof of Theorem sb7f

Step	Hyp	Ref	Expression
1		sb5 2100	. . 3 ⊢ ([z / x]φ ↔ ∃x(x = z ∧ φ))
2	1	sbbii 1653	. 2 ⊢ ([y / z][z / x]φ ↔ [y / z]∃x(x = z ∧ φ))
3		sb7f.1	. . 3 ⊢ Ⅎzφ
4	3	sbco2 2086	. 2 ⊢ ([y / z][z / x]φ ↔ [y / x]φ)
5		sb5 2100	. 2 ⊢ ([y / z]∃x(x = z ∧ φ) ↔ ∃z(z = y ∧ ∃x(x = z ∧ φ)))
6	2, 4, 5	3bitr3i 266	1 ⊢ ([y / x]φ ↔ ∃z(z = y ∧ ∃x(x = z ∧ φ)))

This proof depends on syntax axioms:   ↔ wb 176   ∧ wa 358  ∃wex 1541  Ⅎwnf 1544  [wsb 1648

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is used by:  sb7h  2121