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Theorem sban 1826
Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 3-Feb-2018.)
Assertion
Ref Expression
sban ([y / x](φ ψ) ↔ ([y / x]φ [y / x]ψ))

Proof of Theorem sban
Dummy variable z is distinct from all other variables.
StepHypRef Expression
1 sbanv 1766 . . . 4 ([z / x](φ ψ) ↔ ([z / x]φ [z / x]ψ))
21sbbii 1645 . . 3 ([y / z][z / x](φ ψ) ↔ [y / z]([z / x]φ [z / x]ψ))
3 sbanv 1766 . . 3 ([y / z]([z / x]φ [z / x]ψ) ↔ ([y / z][z / x]φ [y / z][z / x]ψ))
42, 3bitri 173 . 2 ([y / z][z / x](φ ψ) ↔ ([y / z][z / x]φ [y / z][z / x]ψ))
5 ax-17 1416 . . 3 ((φ ψ) → z(φ ψ))
65sbco2v 1818 . 2 ([y / z][z / x](φ ψ) ↔ [y / x](φ ψ))
7 ax-17 1416 . . . 4 (φzφ)
87sbco2v 1818 . . 3 ([y / z][z / x]φ ↔ [y / x]φ)
9 ax-17 1416 . . . 4 (ψzψ)
109sbco2v 1818 . . 3 ([y / z][z / x]ψ ↔ [y / x]ψ)
118, 10anbi12i 433 . 2 (([y / z][z / x]φ [y / z][z / x]ψ) ↔ ([y / x]φ [y / x]ψ))
124, 6, 113bitr3i 199 1 ([y / x](φ ψ) ↔ ([y / x]φ [y / x]ψ))
Colors of variables: wff set class
Syntax hints:   wa 97  wb 98  [wsb 1642
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-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425
This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643
This theorem is referenced by:  sb3an  1829  sbbi  1830  sbmo  1956  moanim  1971  sbabel  2200  nfrexdya  2353  cbvreu  2525  sbcan  2799  sbcang  2800  rmo3  2843  inab  3199  difab  3200  exss  3954  inopab  4411  bdcriota  9318
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