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Theorem sb4bor 1828
Description: Simplified definition of substitution when variables are distinct, expressed via disjunction. (Contributed by Jim Kingdon, 18-Mar-2018.)
Assertion
Ref Expression
sb4bor (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))

Proof of Theorem sb4bor
StepHypRef Expression
1 sb4or 1826 . 2 (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
2 sb2 1760 . . . . 5 (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
3 df-bi 116 . . . . . 6 ((([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)) → (([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) ∧ (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))) ∧ ((([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) ∧ (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))))
43simpri 112 . . . . 5 ((([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) ∧ (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
52, 4mpan2 423 . . . 4 (([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
65alimi 1448 . . 3 (∀𝑥([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) → ∀𝑥([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
76orim2i 756 . 2 ((∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑))) → (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))))
81, 7ax-mp 5 1 (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wb 104  wo 703  wal 1346  [wsb 1755
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528
This theorem depends on definitions:  df-bi 116  df-nf 1454  df-sb 1756
This theorem is referenced by: (None)
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