Theorem wl-ax11-lem10

Description: We now have prepared everything. The unwanted variable u is just in one place left. pm2.61 can be used in conjunction with wl-ax11-lem9 to eliminate the second antecedent. Missing is something along the lines of ax-6 , so we could remove the first antecedent. But the Metamath axioms cannot accomplish this. Such a rule must reside one abstraction level higher than all others: It says that a distinctor implies a distinct variable condition on its contained setvar. This is only needed if such conditions are required, as ax-11v does. The result of this study is for me, that you cannot introduce a setvar capturing this condition, and hope to eliminate it later. (Contributed by Wolf Lammen, 30-Jun-2019)

	Ref	Expression
Assertion	wl-ax11-lem10	$$⊢\forall yy=u\to \left(\neg \forall xx=y\to \left(\forall y\forall x\phi \to \forall x\forall y\phi \right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		wl-ax11-lem8	$$⊢\left(\forall uu=y\wedge \neg \forall xx=y\right)\to \left(\forall u\forall x\left[u/y\right]\phi \leftrightarrow \forall y\forall x\phi \right)$$
2		wl-ax11-lem6	$$⊢\left(\forall uu=y\wedge \neg \forall xx=y\right)\to \left(\forall u\forall x\left[u/y\right]\phi \leftrightarrow \forall x\forall y\phi \right)$$
3	1 2	bitr3d	$$⊢\left(\forall uu=y\wedge \neg \forall xx=y\right)\to \left(\forall y\forall x\phi \leftrightarrow \forall x\forall y\phi \right)$$
4	3	biimpd	$$⊢\left(\forall uu=y\wedge \neg \forall xx=y\right)\to \left(\forall y\forall x\phi \to \forall x\forall y\phi \right)$$
5	4	ex	$$⊢\forall uu=y\to \left(\neg \forall xx=y\to \left(\forall y\forall x\phi \to \forall x\forall y\phi \right)\right)$$
6	5	aecoms	$$⊢\forall yy=u\to \left(\neg \forall xx=y\to \left(\forall y\forall x\phi \to \forall x\forall y\phi \right)\right)$$