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Theorem hbim 1419
Description: If x is not free in φ and ψ, it is not free in (φψ). (Contributed by NM, 5-Aug-1993.) (Proof shortened by O'Cat, 3-Mar-2008.) (Revised by Mario Carneiro, 2-Feb-2015.)
Hypotheses
Ref Expression
hb.1 (φxφ)
hb.2 (ψxψ)
Assertion
Ref Expression
hbim ((φψ) → x(φψ))

Proof of Theorem hbim
StepHypRef Expression
1 ax-4 1381 . . 3 (xφφ)
2 hb.2 . . 3 (ψxψ)
31, 2imim12i 53 . 2 ((φψ) → (xφxψ))
4 ax-i5r 1410 . 2 ((xφxψ) → x(xφψ))
5 hb.1 . . . 4 (φxφ)
65imim1i 54 . . 3 ((xφψ) → (φψ))
76alimi 1324 . 2 (x(xφψ) → x(φψ))
83, 4, 73syl 17 1 ((φψ) → x(φψ))
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1226
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-5 1316  ax-gen 1318  ax-4 1381  ax-i5r 1410
This theorem is referenced by:  hbbi  1422  hbia1  1426  19.21h  1431  19.38  1548  hbsbv  1799  hbmo1  1920  hbmo  1921  moexexdc  1966  2eu4  1975  cleqh  2119  hbral  2331
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