Theorem sbidm 2085

Description: An idempotent law for substitution. (Contributed by NM, 30-Jun-1994.) (Proof shortened by Andrew Salmon, 25-May-2011.)

Assertion

Ref	Expression
sbidm	⊢ ([y / x][y / x]φ ↔ [y / x]φ)

Proof of Theorem sbidm

Step	Hyp	Ref	Expression
1		equsb2 2035	. . 3 ⊢ [y / x]y = x
2		sbequ12r 1920	. . . 4 ⊢ (y = x → ([y / x]φ ↔ φ))
3	2	sbimi 1652	. . 3 ⊢ ([y / x]y = x → [y / x]([y / x]φ ↔ φ))
4	1, 3	ax-mp 5	. 2 ⊢ [y / x]([y / x]φ ↔ φ)
5		sbbi 2071	. 2 ⊢ ([y / x]([y / x]φ ↔ φ) ↔ ([y / x][y / x]φ ↔ [y / x]φ))
6	4, 5	mpbi 199	1 ⊢ ([y / x][y / x]φ ↔ [y / x]φ)

This proof depends on syntax axioms:   ↔ wb 176  [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)