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Theorem termination-RED1
¬ hasS(t) ⊦ 〈RED1(t; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t))
Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by:
termination-RED
,
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 3
Proof steps: 126
Used lemmas:
decl-04
,
decl-01
,
Used spec theorems:
Used simplifier rules:
value
/
val-K
value
/
val-K-ap
nat
/
l-num-01
nat
/
le-add
nat
/
add-num-03
term
/
disj-02
value
/
val-other-ap
term
/
disj-03
term
/
sel-01
term
/
sel-02
term_or_none
/
elim
nat-basic1
/
f
nat-basic2
/
l
term
/
test-02
term
/
disj-01
termfuns
/
hasS-S
nat
/
lep-02
nat
/
le-num-01
nat
/
lels
nat-mult
/
mult-c
nat-mult
/
mult-a
nat-mult
/
one
nat-mult
/
numeral-21
nat
/
l-num-02
nat-mult
/
two-02
termfuns
/
hasS-K
term
/
size-02
nat-basic2
/
z
nat-basic2
/
l-02
nat-basic2
/
c
nat-basic2
/
a
nat-basic2
/
s
nat-basic2
/
succ
term
/
size-03
termfuns
/
hasS-A
term
/
test-03
term
/
elim
termfuns
/
size-some
termfuns
/
hasS-some
term_or_none
/
disj