ivy/test/tc1.ivy

210 строки
7.0 KiB
XML

#lang ivy1.7
# This is an example of an ADT for doing proofs about directed trees
# using the transitive closure encoding. It isn't really complete, but
# gives some idea how we can use manual instantiations in cases where
# we have to go outside the deciable theory. In this case, we need
# some manual proof effort because the definition of transitive
# closure is not expressible in first-order logic. We also wind up
# with one funciton cycle to deal with. Remarkably, only two
# manual proofs are needed and they are ony one line each.
module tree_tc(t) = {
# This module provides a simple interface for manipulating
# directed trees. It has methods for adding an edge, testing
# the successor relation, and determining whcih one node reaches
# anopther (in other words the decendant relation).
action add(x:t,y:t)
action succ(x:t,y:t) returns (r:bool)
action reaches(x:t,y:t) returns (r:bool)
specification {
# The specification is written entirely in terms of `tc`,
# which tracks the transitive closure of the successor
# relation.
relation tc(X:t,Y:t)
# Since initially, the tree is empty, initially tc is empty.
after init {
tc(X,Y) := false
}
# We require that adding an edge preserve the shape as an
# oriented tree. The transitive closure is updated so that x
# and everything that previously reached x now reaches y and
# everything y previously reached. This is the usualy encoding
# of the tree add operations using transitive closure.
around add {
require x ~= y; # no trivial cycles added
require ~tc(y,x); # no non-trivial cycles added
require ~(tc(Z,x) & tc(Z,y)); # no reconvergence added
require ~(tc(x,Z) & tc(Z,y)); # no reconvergence added
...
tc(X,Y) := tc(X,Y) | (X = x | tc(X,x)) & (Y = y | tc(y,Y))
}
# The successor relation has to be computed indirectly using
# from tc. That is, y is a successor of x if x reaches y and
# there is nothing else between x and y (that is, the
# successor relation is the set of non-composite edges of tc).
# We write the two direction of the iff separately, since one
# direction requires additional proof. That is, we need to
# know that every edge in tc is either a tree edge, or a
# composition of two other edges. This is the 'pre fixed
# point' property of transitive closure, which you can see
# below. Since it has a function cycle, we have to manually
# instantiate it here.
after succ {
ensure r -> (tc(x,y) & ~(tc(x,Z) & tc(Z,y)));
ensure (tc(x,y) & ~exists Z. tc(x,Z) & tc(Z,y)) -> r
proof {
assume pre_fp with X=x,Y=y
}
}
# In this encoding, computing the `reaches` relation is
# trivial.
after reaches {
ensure r = tc(x,y)
}
# These invariants state that tc is the transitive closure of
# a tree. Treeness means there are no cycles, and the
# ancestors of every node are totally ordered.
invariant ~tc(X,X)
invariant tc(X,Y) & tc(Z,Y) -> (X = Z | tc(X,Z) | tc(Z,X))
}
implementation {
# The implementation uses the actual tree relation, of which
# `tc` above is the transitive closure.
relation tree(X:t,Y:t)
# We want to define a relation ttc as the true transitive
# closure of 'tree'. A nice way to do this is as a least fixed
# point, essentially a Horn definition. Such a definition might
# look like this if we had such a syntax:
#
# lfp ttc(X,Y) = tree(X,Y) | exists Z. (ttc(X,Z) & ttc(Z,Y))
#
# However, since Horn definitions aren't implemented yet,
# we'll just write here the axioms that the Horn definition
# would generate.
#
# We declare a relation ttc.
relation ttc(X:t,Y:t)
# This axiom says that ttc is a post fixed point
axiom (tree(X,Y) | ttc(X,Z) & ttc(Z,Y)) -> ttc(X,Y)
# This schema says that ttc is a pre fixed point. It is a
# schema because of the forbidden quantifier alternation.
# Essentialy, for any arc in `ttc`, this lets us split cases
# on whether it is tree edge or a composite edge.
schema pre_fp = {
property ttc(X,Y) -> (tree(X,Y) | exists Z. (ttc(X,Z) & ttc(Z,Y)))
}
# This schema says that ttc is the *least* fixed point. That
# is for any fixed point r, ttc is stronger than r. We have to
# state this is a schema and instantiate it manually because
# it isn't first order.
schema lep = {
relation r(X:t,Y:t)
property forall X,Y. ((tree(X,Y) | exists Z. (r(X,Z) & r(Z,Y))) -> r(X,Y))
property ttc(X,Y) -> r(X,Y)
}
# With `ttc` defined in terms of `tree`, we can now define our
# interface implementations. Notice that `reaches` here uses
# `ttc`, which is not computable. For a real implementation,
# we would need to use, say, DFS.
after init {
tree(X,Y) := false;
}
implement add {
tree(x,y) := true;
}
implement succ {
r := tree(x,y);
}
implement reaches {
r := ttc(x,y);
}
}
# The representation invariant is that the abstract `tc` is equal
# to the true transitive closure of `tree`, and that `tree` is the
# set of non-composite edges in tc.
private {
# We state the two directions of equivalence as separate invariants,
# since the proofs are different.
# The tc -> ttc direction only needs the fact that tc is a
# post fixed point. Because this has no quantifier
# alternation, Z3 can get the proof without any help.
invariant [fwd] tc(X,Y) -> ttc(X,Y)
# For ttc -> tc, we need the fact the ttc is the *least* fixed
# point. Notice that this invariant is an instance of the lep
# schema. All we need to say is 'apply lep'. Ivy will infer
# the match r(X,Y) = tc(X,Y). This leaves the proof goal that
# tc is a post fixed point, which Z3 can prove. Have a look at
# the output to see the proof goal Ivy generates.
invariant [rev] ttc(X,Y) -> tc(X,Y)
proof {
apply lep; showgoals
}
# There is actually a subtle issue in the above proof. That
# is, we have to show that `rev` is established initially and
# is preserved by every exported action. We might want a
# separate proof in every case, but here the same argument
# works for all cases. Right now there is not a machanism to
# provide a different proof in every case.
# With the above invariant, Z3 can prove that 'tree' is the
# set of non-composite edges of tc. It suffices to show that
# no tree edge is a composite edge in `tc`. The other
# direction is a consequence the fact that `tc` is a post
# fixed point, so we need not prove it as an invariant.
invariant ~(tree(X,Y) & tc(X,Z) & tc(Z,Y))
}
isolate iso = this
}
type t
instance tr : tree_tc(t)
export tr.add
export tr.succ
export tr.reaches