Theorem sb5f 2040

Description: Equivalence for substitution when y is not free in φ. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
sb6f.1	⊢ Ⅎyφ

Assertion

Ref	Expression
sb5f	⊢ ([y / x]φ ↔ ∃x(x = y ∧ φ))

Proof of Theorem sb5f

Step	Hyp	Ref	Expression
1		sb6f.1	. . 3 ⊢ Ⅎyφ
2	1	sb6f 2039	. 2 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))
3	1	equs45f 1989	. 2 ⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))
4	2, 3	bitr4i 243	1 ⊢ ([y / x]φ ↔ ∃x(x = y ∧ φ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃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-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is used by: (None)