Metamath Proof Explorer


Theorem ifcomnan

Description: Commute the conditions in two nested conditionals if both conditions are not simultaneously true. (Contributed by SO, 15-Jul-2018)

Ref Expression
Assertion ifcomnan ¬φψifφAifψBC=ifψBifφAC

Proof

Step Hyp Ref Expression
1 pm3.13 ¬φψ¬φ¬ψ
2 iffalse ¬φifφAifψBC=ifψBC
3 iffalse ¬φifφAC=C
4 3 ifeq2d ¬φifψBifφAC=ifψBC
5 2 4 eqtr4d ¬φifφAifψBC=ifψBifφAC
6 iffalse ¬ψifψBC=C
7 6 ifeq2d ¬ψifφAifψBC=ifφAC
8 iffalse ¬ψifψBifφAC=ifφAC
9 7 8 eqtr4d ¬ψifφAifψBC=ifψBifφAC
10 5 9 jaoi ¬φ¬ψifφAifψBC=ifψBifφAC
11 1 10 syl ¬φψifφAifψBC=ifψBifφAC