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Theorem xorbi1d 1256
Description: Deduction joining an equivalence and a right operand to form equivalence of exclusive-or. (Contributed by Jim Kingdon, 7-Oct-2018.)
Hypothesis
Ref Expression
xorbid.1 (φ → (ψχ))
Assertion
Ref Expression
xorbi1d (φ → ((ψθ) ↔ (χθ)))

Proof of Theorem xorbi1d
StepHypRef Expression
1 xorbid.1 . . . 4 (φ → (ψχ))
21orbi1d 692 . . 3 (φ → ((ψ θ) ↔ (χ θ)))
31anbi1d 441 . . . 4 (φ → ((ψ θ) ↔ (χ θ)))
43notbid 579 . . 3 (φ → (¬ (ψ θ) ↔ ¬ (χ θ)))
52, 4anbi12d 445 . 2 (φ → (((ψ θ) ¬ (ψ θ)) ↔ ((χ θ) ¬ (χ θ))))
6 df-xor 1252 . 2 ((ψθ) ↔ ((ψ θ) ¬ (ψ θ)))
7 df-xor 1252 . 2 ((χθ) ↔ ((χ θ) ¬ (χ θ)))
85, 6, 73bitr4g 212 1 (φ → ((ψθ) ↔ (χθ)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4   wa 97  wb 98   wo 616  wxo 1251
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-in1 532  ax-in2 533  ax-io 617
This theorem depends on definitions:  df-bi 110  df-xor 1252
This theorem is referenced by:  xorbi12d  1257
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