Theorem eq6s 827

Description: Rule that applies eq6 826 to antecedent.

Hypothesis

Ref	Expression
eq6s.1	⊢ (∀z ¬ ∀x x = y → φ)

Assertion

Ref	Expression
eq6s	⊢ (¬ ∀x x = y → φ)

Proof of Theorem eq6s

Step	Hyp	Ref	Expression
1		eq6 826	. 2 ⊢ (¬ ∀x x = y → ∀z ¬ ∀x x = y)
2		eq6s.1	. 2 ⊢ (∀z ¬ ∀x x = y → φ)
3	1, 2	syl 12	1 ⊢ (¬ ∀x x = y → φ)

Syntax hints:  ¬ wn 1   → wi 2  ∀wal 672   = weq 797

This theorem is referenced by:  sb9i 920  sbal1 996  sbal2 1005  ralcom2 1314

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679

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