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Theorem sb4bor 1835
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 1833 . 2 (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
2 sb2 1767 . . . . 5 (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
3 df-bi 117 . . . . . 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 113 . . . . 5 ((([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) ∧ (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
52, 4mpan2 425 . . . 4 (([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
65alimi 1455 . . 3 (∀𝑥([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) → ∀𝑥([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
76orim2i 761 . 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 104  wb 105  wo 708  wal 1351  [wsb 1762
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-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535
This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763
This theorem is referenced by: (None)
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