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./test/should-error/all_intro_advice_partial.pf:6.3-6.4: incomplete proof | ||
Goal: | ||
(all y:bool, z:bool, w:bool. (all n:bool. (x or y or y or not (x)))) | ||
(all y:bool, z:bool, w:bool. (all n:bool. (x or y or y or not x))) | ||
Advice: | ||
You can complete the proof with: | ||
arbitrary y:bool | ||
followed by a proof of: | ||
(all z:bool, w:bool. (all n:bool. (x or y or y or not (x)))) | ||
(all z:bool, w:bool. (all n:bool. (x or y or y or not x))) | ||
Givens: | ||
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./test/should-error/prefix_operator.pf:4.3-4.4: incomplete proof | ||
Goal: | ||
not false | ||
Advice: | ||
You can complete the proof with: | ||
assume label: false | ||
followed by a proof of: | ||
false | ||
Givens: | ||
|
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import List | ||
import Nat | ||
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function sum(List<Nat>) -> Nat { | ||
sum(empty) = 0 | ||
sum(node(n, ns)) = n + sum(ns) | ||
} | ||
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theorem foldr_eq_sum : all ls : List<Nat>. sum(ls) = foldr(ls, 0, operator+) | ||
proof | ||
induction List<Nat> | ||
case empty { | ||
suffices ? by definition sum | ||
? | ||
} | ||
case node(a, d) { | ||
? | ||
} | ||
end |
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./test/should-error/sum_foldr.pf:14.5-14.33: | ||
suffices to prove: | ||
0 = foldr([], 0, operator +) | ||
./test/should-error/sum_foldr.pf:15.5-15.6: incomplete proof | ||
Goal: | ||
0 = foldr([], 0, operator +) | ||
Advice: | ||
To prove this equality, one of these statements might help: | ||
definition | ||
rewrite | ||
equations | ||
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Givens: | ||
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