Metamath Proof Explorer


Theorem ifeq3da

Description: Given an expression C containing if ( ps , E , F ) , substitute (hypotheses .1 and .2) and evaluate (hypotheses .3 and .4) it for both cases at the same time. (Contributed by Thierry Arnoux, 13-Dec-2021)

Ref Expression
Hypotheses ifeq3da.1 ifψEF=EC=G
ifeq3da.2 ifψEF=FC=H
ifeq3da.3 φG=A
ifeq3da.4 φH=B
Assertion ifeq3da φifψAB=C

Proof

Step Hyp Ref Expression
1 ifeq3da.1 ifψEF=EC=G
2 ifeq3da.2 ifψEF=FC=H
3 ifeq3da.3 φG=A
4 ifeq3da.4 φH=B
5 iftrue ψifψEF=E
6 5 1 syl ψC=G
7 6 adantl φψC=G
8 3 adantr φψG=A
9 7 8 eqtr2d φψA=C
10 iffalse ¬ψifψEF=F
11 10 2 syl ¬ψC=H
12 11 adantl φ¬ψC=H
13 4 adantr φ¬ψH=B
14 12 13 eqtr2d φ¬ψB=C
15 9 14 ifeqda φifψAB=C