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42 | 42 | For the inductive step, assume that $t = \Atom{f}{t_1, \dots, t_k}$
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43 | 43 | and that the claim holds for $t_1$, \dots, $t_k$. Then
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44 | 44 | \begin{align*}
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45 |
| - \Value{t}{M}[s_1] & = \Value{\Atom{f}{t_1, \dots, t_k}}{M}[s_1] =\\ |
46 |
| - & = \Assign{f}{M}(\Value{t_1}{M}[s_1], \dots, \Value{t_k}{M}[s_1]) |
| 45 | + \Value{t}{M}[s_1] & = \Value{\Atom{f}{t_1, \dots, t_k}}{M}[s_1] \\ |
| 46 | + & = \Assign{f}{M}(\Value{t_1}{M}[s_1], \dots, \Value{t_k}{M}[s_1]). |
47 | 47 | \intertext{For $j = 1$, \dots,~$k$, the !!{variable}s of~$t_j$ are
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48 |
| - among $x_1$, \dots,~$x_n$. By induction hypothesis, |
| 48 | + among $x_1$, \dots,~$x_n$. By the induction hypothesis, |
49 | 49 | $\Value{t_j}{M}[s_1] = \Value{t_j}{M}[s_2]$. So,}
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50 |
| -\Value{t}{M}[s_1] & = \Value{\Atom{f}{t_1, \dots, t_k}}{M}[s_1] =\\ |
51 |
| - & = \Assign{f}{M}(\Value{t_1}{M}[s_1], \dots, \Value{t_k}{M}[s_1]) =\\ |
52 |
| - & = \Assign{f}{M}(\Value{t_1}{M}[s_2], \dots, \Value{t_k}{M}[s_2]) =\\ |
| 50 | +\Value{t}{M}[s_1] & = \Value{\Atom{f}{t_1, \dots, t_k}}{M}[s_1] \\ |
| 51 | + & = \Assign{f}{M}(\Value{t_1}{M}[s_1], \dots, \Value{t_k}{M}[s_1]) \\ |
| 52 | + & = \Assign{f}{M}(\Value{t_1}{M}[s_2], \dots, \Value{t_k}{M}[s_2]) \\ |
53 | 53 | & = \Value{\Atom{f}{t_1, \dots, t_k}}{M}[s_2] = \Value{t}{M}[s_2].
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54 | 54 | \end{align*}
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55 | 55 | \end{proof}
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