Nuprl Rule : dependentIntersectionEquality
This rule proved as lemma rule_free_from_atom_equality_true3 in file rules/rules_free_from_atom.v
at https://github.com/vrahli/NuprlInCoq
H ⊢ x1:a1 ⋂ b1 = x2:a2 ⋂ b2 ∈ Type
BY dependentIntersectionEquality y
H ⊢ a1 = a2 ∈ Type
H y:a1 ⊢ b1[y/x1] = b2[y/x2] ∈ Type
Definitions occuring in rule :
dep-isect: x:A ⋂ B[x],
equal: s = t ∈ T,
universe: Type,
axiom: Ax
Latex:
H \mvdash{} x1:a1 \mcap{} b1 = x2:a2 \mcap{} b2
BY dependentIntersectionEquality y
H \mvdash{} a1 = a2
H y:a1 \mvdash{} b1[y/x1] = b2[y/x2]
Date html generated:
2019_06_20-PM-04_12_16
Last ObjectModification:
2015_11_23-PM-03_41_59
Theory : rules
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