, We then define M n = (W n , ? n , V n ). If moreover we have that S(w) ? S(v) implies w ? v, then we say that M is an expanding model. We define stratified and expanding posets similarly
, ? v 2 , v 1 ? S k1 (w 1 ) and v 1 Z i v 2 , and 2. for all j 2 ? [0, k 2 ) there exist j 1 ? [0, k 1 ) and (u 1 , u 2 ) ? W 1 × W 2 such that u 1 ? S j1
, ) and v 1 Z i v 2 , and 2. for all j 1 ? [0, k 1 ) there exist j 2 ? [0, k 2 ) and (u 1 , u 2 ) ? W 1 × W 2 such that u 2 ? S j2, S k1 (w 1 ) ? v 1 , v 2 ? S k2
, As was the case before, the following lemma states that two bounded Ubisimilar models agree on small-enough L U formulas
, Lemma 23. Given two ITL e models M 1 and M 2 and a bounded U-bisimulation Z n ? ? ?Z 0 between them, for all m ? n and (w 1 , w 2 ) ? W 1 × W 2 , if w 1 Z m w 2 then for all ? ? L U such that ? ? m, M 1
, We only consider the new case, where ? = ? U ?. From left to right, assume that M 1 , w 1 ? ? U ?. Then, there exists i 1 ? 0 such that M 1 , S i1 (w 1 ) ? ? and for all j 1 satisfying 0 ? j 1 < i 1 , M 1 , S j1 (w 1 ) ? ?. By Forth U, there exist i 2 ? 0 and (v 1 , v 2 ) ? W 1 × W 2 such that 1, L U such that ? ? m
, Since ? ? m ? 1, it follows from the induction hypothesis that M 2 , v 2 ? ?, and by ?-monotonicity, M 2 , S i2 (w 2 ) ? ?. Now take any j 2 satisfying 0 ? j 2 < i 2 . Using (2), the fact that ? ? m ? 1, and the induction hypothesis, M 1 , S i1 (w 1 ) ? ?, by ?-monotonicity we see that M 1 , v 1 ? ?
, B is initial if and only if b ? n ? m ? ?(a)
, is terminal if and only if b ? (n ? m ? ?(a), n ? m + 1], and 3. B is regular if and only if b > n ? m + 1
, But then if b ? n ? ?(a) ? m we have that V (a2 m , b) = ? as well, so that B is initial if and only if b ? n ? ?(a) ? m, while if b ? (n ? m ? ?(a), n ? m + 1] it is neither initial nor regular, hence it is terminal, Proof. Let B = B m (a, b) be any block. First observe that ?(a2 m ) = ?(a)+m ? m, so that p ? V
, Then, if x ? W is m-initial, Lemma 29. Let n ? 0 and M n =
, Let x = (i, b) and a be such that x ? B = B m (a, b), p.28
, we see by Lemma 28.2 that B ? is terminal, and as b ? b ? that y = (i, b ? ) ? x, as needed
, If B and B ? are regular m-blocks then B and B ? are congruent if and only if they have the same height
, Since b > n ? m + 1 and ?(c2 m ) = ?(c) + m > m for c ? {a, a ? }, we see that b > n ? ?(a2 m ) and also b > n ? ?(a ? 2 m ), so that p ? V (a2 m , b) ? V (a ? 2 m , b ? ). We conclude that for all k ? [0, 2 m ], p ? (a ? 1)2 m + k, b if and only if p ? (a ? ? 1)2 m + k, b , i.e. B and B ? are congruent. If instead b = b ? , assume without loss of generality that b < b ? . Since b > n?m+ 1 we have that k ?= 2 n?b ? [1, 2 m ), Suppose that B = B m (a, b) and B ? = B m (a ? , b ? ) are regular, so that by Lemma 28.3, b, b ? > n ? m + 1. If b = b ? and k ? [1, 2 m ) then by Lemma 26, ? (a ? 1k, b
, If B and B ? are congruent m-blocks, then: 1. the first halves of B and B ? are congruent, 2. the second halves of B and B ? are congruent, and 3. the successors of the second halves of B and B ? are congruent
, For the third item, the congruence of the successors of the second halves of B and B ? is shown by a case-by-case analysis: if B and B ? are regular, then the successors of the second halves are either both terminal or both regular with the same height. If B, B ? are not regular
, The relations (? m ) m<n form a graded R-bisimulation on M n
, Note that ? m is symmetric, so we only check the 'forth' clauses. Below, assume that x = (x 1 , x 2 ), y = (y 1 , y 2 ) and x ? m
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