Nuprl Rule : functionEquality
This rule proved as lemma rule_function_equality_true in file rules/rules_function.v
at https://github.com/vrahli/NuprlInCoq
H ⊢ (x1:a1 ⟶ b1) = (x2:a2 ⟶ b2) ∈ Type
BY functionEquality y
H ⊢ a1 = a2 ∈ Type
H y:a1 ⊢ b1[y/x1] = b2[y/x2] ∈ Type
Definitions occuring in rule :
function: x:A ⟶ B[x],
equal: s = t ∈ T,
universe: Type,
axiom: Ax
Latex:
H \mvdash{} (x1:a1 {}\mrightarrow{} b1) = (x2:a2 {}\mrightarrow{} b2)
BY functionEquality y
H \mvdash{} a1 = a2
H y:a1 \mvdash{} b1[y/x1] = b2[y/x2]
Date html generated:
2019_06_20-PM-04_11_48
Last ObjectModification:
2016_07_08-PM-03_48_57
Theory : rules
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