Theorem 4atexlemnslpq

Description: Lemma for 4atexlem7 . (Contributed by NM, 23-Nov-2012)

	Ref	Expression
Hypothesis	4thatlem.ph	$$⊢\phi \leftrightarrow \left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(S\in A\wedge \left(R\in A\wedge \neg R\underset{\dot{}}{\le }W\wedge P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\wedge \left(T\in A\wedge U\underset{\dot{}}{\vee }T=V\underset{\dot{}}{\vee }T\right)\right)\wedge \left(P\ne Q\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)$$
Assertion	4atexlemnslpq	$$⊢\phi \to \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)$$

Proof

Step	Hyp	Ref	Expression
1		4thatlem.ph	$$⊢\phi \leftrightarrow \left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(S\in A\wedge \left(R\in A\wedge \neg R\underset{\dot{}}{\le }W\wedge P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\wedge \left(T\in A\wedge U\underset{\dot{}}{\vee }T=V\underset{\dot{}}{\vee }T\right)\right)\wedge \left(P\ne Q\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)$$
2		simp3r	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(S\in A\wedge \left(R\in A\wedge \neg R\underset{\dot{}}{\le }W\wedge P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\wedge \left(T\in A\wedge U\underset{\dot{}}{\vee }T=V\underset{\dot{}}{\vee }T\right)\right)\wedge \left(P\ne Q\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)\to \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)$$
3	1 2	sylbi	$$⊢\phi \to \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)$$