Nuprl Rule : setEquality
This rule proved as lemma rule_set_equality_true in file rules/rules_set.v at https://github.com/vrahli/NuprlInCoq
H ⊢ {x1:A1| B1} = {x2:A2| B2} ∈ Type
BY setEquality x
H ⊢ A1 = A2 ∈ Type
H x:A1 ⊢ B1[x/x1] = B2[x/x2] ∈ Type
Definitions occuring in rule :
set: {x:A| B[x]} ,
equal: s = t ∈ T,
universe: Type,
axiom: Ax
Latex:
H \mvdash{} \{x1:A1| B1\} = \{x2:A2| B2\}
BY setEquality x
H \mvdash{} A1 = A2
H x:A1 \mvdash{} B1[x/x1] = B2[x/x2]
Date html generated:
2019_06_20-PM-04_11_48
Last ObjectModification:
2016_07_08-PM-03_48_48
Theory : rules
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