Theorem sb6a 2004

Description: Equivalence for substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sb6a	⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → [𝑥 / 𝑦]𝜑))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem sb6a

Step	Hyp	Ref	Expression
1		sb6 1898	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
2		sbequ12 1782	. . . . 5 ⊢ (𝑦 = 𝑥 → (𝜑 ↔ [𝑥 / 𝑦]𝜑))
3	2	equcoms 1719	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑥 / 𝑦]𝜑))
4	3	pm5.74i 180	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → [𝑥 / 𝑦]𝜑))
5	4	albii 1481	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → [𝑥 / 𝑦]𝜑))
6	1, 5	bitri 184	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → [𝑥 / 𝑦]𝜑))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1362  [wsb 1773

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545

This theorem depends on definitions:  df-bi 117  df-sb 1774

This theorem is referenced by: (None)