Theorems · Theorem · commutative algebra
WithVal.equivWithVal_symm
Deprecated since 2026-01-27Use WithVal.congr_symm instead.
∀ {R : Type u_1} {Γ₀ : Type u_2} [inst : LinearOrderedCommGroupWithZero Γ₀] [inst_1 : Ring R] {Γ'₀ : Type u_3}
[inst_2 : LinearOrderedCommGroupWithZero Γ'₀] (v : Valuation R Γ₀) (w : Valuation R Γ'₀),
(WithVal.congr v w (RingEquiv.refl R)).symm = WithVal.congr w v (RingEquiv.refl R)- Defined in
- Mathlib.Topology.Algebra.Valued.WithVal
- Cited by
- 0 results in Mathlib
- Foundations
- Depth 28 from the axioms · uses propext, Quot.sound
Around this declaration
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Cites8
Mathlib declarations this one mentions in its statement or cites explicitly in its proof. Plumbing is filtered out.
- Ringstatement and proof · cited by 7,463
- RingEquivstatement · cited by 1,147
- Valuationstatement and proof · cited by 823
- RingEquiv.symmstatement · cited by 567
- LinearOrderedCommGroupWithZerostatement and proof · cited by 528
- WithValstatement · cited by 151
- RingEquiv.reflstatement · cited by 72
- WithVal.congrstatement · cited by 10
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