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Theorem sb6a 1880
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
StepHypRef Expression
1 sb6 1782 . 2 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
2 sbequ12 1670 . . . . 5 (𝑦 = 𝑥 → (𝜑 ↔ [𝑥 / 𝑦]𝜑))
32equcoms 1610 . . . 4 (𝑥 = 𝑦 → (𝜑 ↔ [𝑥 / 𝑦]𝜑))
43pm5.74i 173 . . 3 ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑦 → [𝑥 / 𝑦]𝜑))
54albii 1375 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → [𝑥 / 𝑦]𝜑))
61, 5bitri 177 1 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → [𝑥 / 𝑦]𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 102  wal 1257  [wsb 1661
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-11 1413  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-sb 1662
This theorem is referenced by: (None)
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