Nuprl Lemma : assert-pushdownC-test
∀a,b,c:ℤ.
((↑(((a =z c) ∧b c <z b) ∧b (¬b((a =z b) ∨b(b + 2 =z c)))))
⇒ {(¬(a = b ∈ ℤ)) ∧ (¬((b + 2) = c ∈ ℤ)) ∧ c < b ∧ (a = c ∈ ℤ)})
Proof
Definitions occuring in Statement :
bor: p ∨bq,
band: p ∧b q,
bnot: ¬bb,
assert: ↑b,
lt_int: i <z j,
eq_int: (i =z j),
less_than: a < b,
guard: {T},
all: ∀x:A. B[x],
not: ¬A,
implies: P ⇒ Q,
and: P ∧ Q,
add: n + m,
natural_number: $n,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x],
implies: P ⇒ Q,
or: P ∨ Q,
true: True,
member: t ∈ T,
prop: ℙ,
uall: ∀[x:A]. B[x],
exists: ∃x:A. B[x],
uimplies: b supposing a,
sq_type: SQType(T),
guard: {T},
band: p ∧b q,
ifthenelse: if b then t else f fi ,
btrue: tt,
uiff: uiff(P;Q),
and: P ∧ Q,
top: Top,
bnot: ¬bb,
assert: ↑b,
bfalse: ff,
false: False,
bor: p ∨bq,
cand: A c∧ B,
nequal: a ≠ b ∈ T
Lemmas referenced :
true_wf,
eq_int_wf,
bool_cases_sqequal,
subtype_base_sq,
bool_subtype_base,
eqtt_to_assert,
assert_of_eq_int,
lt_int_wf,
bool_wf,
assert_of_lt_int,
testxxx_lemma,
istype-void,
eqff_to_assert,
assert-bnot,
neg_assert_of_eq_int,
istype-assert,
bool_cases,
band_wf,
btrue_wf,
bfalse_wf,
bnot_wf,
bor_wf,
istype-int
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :lambdaFormation_alt,
cut,
Error :inlFormation_alt,
natural_numberEquality,
Error :universeIsType,
introduction,
extract_by_obid,
sqequalHypSubstitution,
hypothesis,
unionElimination,
thin,
isectElimination,
hypothesisEquality,
Error :dependent_pairFormation_alt,
equalityTransitivity,
equalitySymmetry,
Error :equalityIstype,
Error :inhabitedIsType,
productElimination,
promote_hyp,
dependent_functionElimination,
instantiate,
because_Cache,
independent_isectElimination,
independent_functionElimination,
sqequalRule,
cumulativity,
Error :isect_memberEquality_alt,
voidElimination,
addEquality,
independent_pairFormation
Latex:
\mforall{}a,b,c:\mBbbZ{}.
((\muparrow{}(((a =\msubz{} c) \mwedge{}\msubb{} c <z b) \mwedge{}\msubb{} (\mneg{}\msubb{}((a =\msubz{} b) \mvee{}\msubb{}(b + 2 =\msubz{} c)))))
{}\mRightarrow{} \{(\mneg{}(a = b)) \mwedge{} (\mneg{}((b + 2) = c)) \mwedge{} c < b \mwedge{} (a = c)\})
Date html generated:
2019_06_20-PM-02_25_27
Last ObjectModification:
2019_01_22-AM-08_31_16
Theory : num_thy_1
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