3 Merge commits $L$ and $R$ using merge base $M$:
5 C \hasparents \{ L, R \}
7 \patchof{C} = \patchof{L}
11 We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
13 This can also be used for dependency re-insertion, by setting
14 $L \in \pn$, $R \in \pry$, $M = \baseof{R}$.
16 \subsection{Conditions}
17 \[ \eqn{ Ingredients }{
23 R \in \py : & \baseof{R} \ge \baseof{L}
24 \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
25 R \in \pn : & M = \baseof{L} \\
26 \text{otherwise} : & \false
29 \[ \eqn{ Merge Acyclic }{
34 \[ \eqn{ Removal Merge Ends }{
35 X \not\haspatch \p \land
39 \pendsof{Y}{\py} = \pendsof{M}{\py}
41 \[ \eqn{ Addition Merge Ends }{
42 X \not\haspatch \p \land
43 M \nothaspatch \p \land
46 \bigforall_{E \in \pendsof{X}{\py}} E \le Y
49 \[ \eqn{ Foreign Merges }{
50 \patchof{L} = \bot \implies \patchof{R} = \bot
53 \subsection{Non-Topbloke merges}
55 We require both $\patchof{L} = \bot$ and $\patchof{R} = \bot$
56 (Foreign Merges, above).
57 I.e. not only is it forbidden to merge into a Topbloke-controlled
58 branch without Topbloke's assistance, it is also forbidden to
59 merge any Topbloke-controlled branch into any plain git branch.
61 Given those conditions, Tip Merge and Merge Acyclic do not apply.
62 By Foreign Contents of $L$, $\patchof{M} = \bot$ as well.
63 So by Foreign Contents for any $A \in \{L,M,R\}$,
64 $\forall_{\p, D \in \py} D \not\le A$
65 so $\pendsof{A}{\py} = \{ \}$ and the RHS of both Merge Ends
66 conditions are satisifed.
68 So a plain git merge of non-Topbloke branches meets the conditions and
69 is therefore consistent with our model.
71 \subsection{No Replay}
73 By definition of $\merge$,
74 $D \isin C \implies D \isin L \lor D \isin R \lor D = C$.
76 Ingredients Prevent Replay applies. $\qed$
78 \subsection{Unique Base}
80 Need to consider only $C \in \py$, ie $L \in \py$,
81 and calculate $\pendsof{C}{\pn}$. So we will consider some
82 putative ancestor $A \in \pn$ and see whether $A \le C$.
84 By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$.
85 But $C \in \py$ and $A \in \pn$ so $A \neq C$.
86 Thus $A \le C \equiv A \le L \lor A \le R$.
88 By Unique Base of L and Transitive Ancestors,
89 $A \le L \equiv A \le \baseof{L}$.
91 \subsubsection{For $R \in \py$:}
93 By Unique Base of $R$ and Transitive Ancestors,
94 $A \le R \equiv A \le \baseof{R}$.
96 But by Tip Merge condition on $\baseof{R}$,
97 $A \le \baseof{L} \implies A \le \baseof{R}$, so
98 $A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
99 Thus $A \le C \equiv A \le \baseof{R}$.
100 That is, $\baseof{C} = \baseof{R}$.
102 \subsubsection{For $R \in \pn$:}
104 By Tip Merge condition and since $M \le R$,
105 $A \le \baseof{L} \implies A \le R$, so
106 $A \le R \lor A \le \baseof{L} \equiv A \le R$.
107 Thus $A \le C \equiv A \le R$.
108 That is, $\baseof{C} = R$.
112 \subsection{Coherence and Patch Inclusion}
114 Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
115 This involves considering $D \in \py$.
117 \subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
118 $D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L
119 \in \py$ ie $\neg[ L \nothaspatch \p ]$ by Tip Own Contents for $L$).
121 Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
123 \subsubsection{For $L \haspatch \p, R \haspatch \p$:}
124 $D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
125 (Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.)
127 Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
129 For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
130 \equiv D \isin L \lor D \isin R$.
131 (Likewise $D \le C \equiv D \le X \lor D \le Y$.)
133 Consider $D \neq C, D \isin X \land D \isin Y$:
134 By $\merge$, $D \isin C$. Also $D \le X$
135 so $D \le C$. OK for $C \haspatch \p$.
137 Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
138 By $\merge$, $D \not\isin C$.
139 And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
140 OK for $C \haspatch \p$.
142 Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
143 $D \not\le X$ so $D \not\le M$ so $D \not\isin M$.
144 Thus by $\merge$, $D \isin C$. And $D \le Y$ so $D \le C$.
145 OK for $C \haspatch \p$.
147 So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
149 \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
151 $M \haspatch \p \implies C \nothaspatch \p$.
152 $M \nothaspatch \p \implies C \haspatch \p$.
156 One of the Merge Ends conditions applies.
157 Recall that we are considering $D \in \py$.
158 $D \isin Y \equiv D \le Y$. $D \not\isin X$.
159 We will show for each of
161 if $M \haspatch \p$, $D \not\isin C$,
162 whereas if $M \nothaspatch \p$, $D \isin C \equiv \land D \le C$.
164 Consider $D = C$: Thus $C \in \py, L \in \py$.
165 By Tip Own Contents, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,
166 therefore we must have $L=Y$, $R=X$.
167 By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
168 by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
171 Consider $D \neq C, M \nothaspatch \p, D \isin Y$:
172 $D \le Y$ so $D \le C$.
173 $D \not\isin M$ so by $\merge$, $D \isin C$. OK.
175 Consider $D \neq C, M \nothaspatch \p, D \not\isin Y$:
176 $D \not\le Y$. If $D \le X$ then
177 $D \in \pancsof{X}{\py}$, so by Addition Merge Ends and
178 Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
179 Thus $D \not\le C$. By $\merge$, $D \not\isin C$. OK.
181 Consider $D \neq C, M \haspatch \p, D \isin Y$:
182 $D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
183 and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
184 Thus $D \isin M$. By $\merge$, $D \not\isin C$. OK.
186 Consider $D \neq C, M \haspatch \p, D \not\isin Y$:
187 By $\merge$, $D \not\isin C$. OK.
191 \subsection{Base Acyclic}
193 This applies when $C \in \pn$.
194 $C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
196 Consider some $D \in \py$.
198 By Base Acyclic of $L$, $D \not\isin L$. By the above, $D \not\isin
199 R$. And $D \neq C$. So $D \not\isin C$.
203 \subsection{Tip Contents}
205 We need worry only about $C \in \py$.
206 And $\patchof{C} = \patchof{L}$
207 so $L \in \py$ so $L \haspatch \p$. We will use the Unique Base
208 of $C$, and its Coherence and Patch Inclusion, as just proved.
210 Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
211 \p$ and by Coherence/Inclusion $C \haspatch \p$ . If $R \not\in \py$
212 then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
213 of $\nothaspatch$, $M \nothaspatch \p$. So by Coherence/Inclusion $C
214 \haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
216 We will consider an arbitrary commit $D$
217 and prove the Exclusive Tip Contents form.
219 \subsubsection{For $D \in \py$:}
220 $C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
223 \subsubsection{For $D \not\in \py, R \not\in \py$:}
225 $D \neq C$. By Tip Contents of $L$,
226 $D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition,
227 $D \isin L \equiv D \isin M$. So by $\merge$, $D \isin
228 C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
229 Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
231 \subsubsection{For $D \not\in \py, R \in \py$:}
236 $D \isin L \equiv D \isin \baseof{L}$ and
237 $D \isin R \equiv D \isin \baseof{R}$.
239 Apply Tip Merge condition.
240 If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
241 Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
242 $\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
243 $D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
245 So $D \isin M \equiv D \isin L$ so by $\merge$,
246 $D \isin C \equiv D \isin R$. But from Unique Base,
247 $\baseof{C} = \baseof{R}$.
248 Therefore $D \isin C \equiv D \isin \baseof{C}$. OK.
252 \subsection{Foreign Inclusion}
254 Consider some $D$ s.t. $\patchof{D} = \bot$.
255 By Foreign Inclusion of $L, M, R$:
256 $D \isin L \equiv D \le L$;
257 $D \isin M \equiv D \le M$;
258 $D \isin R \equiv D \le R$.
260 \subsubsection{For $D = C$:}
262 $D \isin C$ and $D \le C$. OK.
264 \subsubsection{For $D \neq C, D \isin M$:}
266 Thus $D \le M$ so $D \le L$ and $D \le R$ so $D \isin L$ and $D \isin
267 R$. So by $\merge$, $D \isin C$. And $D \le C$. OK.
269 \subsubsection{For $D \neq C, D \not\isin M, D \isin X$:}
271 By $\merge$, $D \isin C$.
272 And $D \isin X$ means $D \le X$ so $D \le C$.
275 \subsubsection{For $D \neq C, D \not\isin M, D \not\isin L, D \not\isin R$:}
277 By $\merge$, $D \not\isin C$.
278 And $D \not\le L, D \not\le R$ so $D \not\le C$.
283 \subsection{Foreign Contents}
285 Only relevant if $\patchof{L} = \bot$, in which case
286 $\patchof{C} = \bot$ and by Foreign Merges $\patchof{R} = \bot$,
287 so Totally Foreign Contents applies. $\qed$