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Theorem 2sb5 1859
 Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.)
Assertion
Ref Expression
2sb5 ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ 𝜑))
Distinct variable groups:   𝑥,𝑦,𝑧   𝑦,𝑤
Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧,𝑤)

Proof of Theorem 2sb5
StepHypRef Expression
1 sb5 1767 . 2 ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥(𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑))
2 19.42v 1786 . . . 4 (∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤𝜑)) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤𝜑)))
3 anass 381 . . . . 5 (((𝑥 = 𝑧𝑦 = 𝑤) ∧ 𝜑) ↔ (𝑥 = 𝑧 ∧ (𝑦 = 𝑤𝜑)))
43exbii 1496 . . . 4 (∃𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ 𝜑) ↔ ∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤𝜑)))
5 sb5 1767 . . . . 5 ([𝑤 / 𝑦]𝜑 ↔ ∃𝑦(𝑦 = 𝑤𝜑))
65anbi2i 430 . . . 4 ((𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤𝜑)))
72, 4, 63bitr4ri 202 . . 3 ((𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ ∃𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ 𝜑))
87exbii 1496 . 2 (∃𝑥(𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ ∃𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ 𝜑))
91, 8bitri 173 1 ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ 𝜑))
 Colors of variables: wff set class Syntax hints:   ∧ wa 97   ↔ wb 98  ∃wex 1381  [wsb 1645 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427 This theorem depends on definitions:  df-bi 110  df-sb 1646 This theorem is referenced by:  opelopabsbALT  3996
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