ifcomnan - Metamath Proof Explorer

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Theorem ifcomnan 4137

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

**Assertion**
Ref	Expression
ifcomnan	$/\$

**Proof of Theorem ifcomnan**
Step	Hyp	Ref	Expression
1		pm3.13 522	. 2 $/\$ $\/$
2		iffalse 4095	. . . 4
3		iffalse 4095	. . . . 5
4	3	ifeq2d 4105	. . . 4
5	2, 4	eqtr4d 2659	. . 3
6		iffalse 4095	. . . . 5
7	6	ifeq2d 4105	. . . 4
8		iffalse 4095	. . . 4
9	7, 8	eqtr4d 2659	. . 3
10	5, 9	jaoi 394	. 2 $\/$
11	1, 10	syl 17	1 $/\$

Colors of variables: wff setvar class

Syntax hints:

wn 3

wi 4 $\/$ wo 383 $/\$ wa 384

wceq 1483

cif 4086

This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1722 ax-4 1737 ax-5 1839 ax-6 1888 ax-7 1935 ax-9 1999 ax-10 2019 ax-11 2034 ax-12 2047 ax-13 2246 ax-ext 2602

This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-tru 1486 df-ex 1705 df-nf 1710 df-sb 1881 df-clab 2609 df-cleq 2615 df-clel 2618 df-nfc 2753 df-rab 2921 df-v 3202 df-un 3579 df-if 4087

This theorem is referenced by: mdetunilem6 20423