In reply to Adarsh24:
// An assertion of the form
// seq |-> expr1 ## [1:$] seq2
// can only fail if expr1 is false. <<<**********
// If expr1==1 then there are an infinite number of possibilities
// for seq2 to be true.
seq |-> expr1 ## [1:$] seq2 // is equivalent to
seq |-> expr1 ##1 seq2 or
expr1 ##2 seq2 or
...
expr1 ##n seq2; // n= $
// The difference between something like
##1 !c[->1]
##1 (reg2[17:1] == 17'h00000) [->1]
##1 (reg2[17:1] == 17'h1FFFF) [->1]
// and this
##[1:$] (!c)
##[1:$] (reg2[17:1] == 17'h00000)
##[1:$] (reg2[17:1] == 17'h1FFFF)
// is that with the goto operator, when
(!c) // occurs
// no other occurrences of !c is further tested.
// For example, the following 3 sequences are possibilities
##2 (!c)
##4 (reg2[17:1] == 17'h00000)
##6 (reg2[17:1] == 17'h1FFFF)
// or ..
##3 (!c)
##5 (reg2[17:1] == 17'h00000)
##6 (reg2[17:1] == 17'h1FFFF)
// In essence, the use of the goto operator it is equivalent to
first_match(##[1:$] (!c))
##0 first_match(##[1:$] (reg2[17:1] == 17'h00000))
##0 first_match(##[1:$] (reg2[17:1] == 17'h1FFFF))
/////////////////////
// Taking the path out, I would write it as
property example1;
disable iff (!(testmode))
@(posedge clk) (a && b) |-> (1, $display ("Checking SEQ1"))
##0 (c && (reg1[16:0] == 17'h0000F))
##1 !c[->1]
##1 (reg2[17:1] == 17'h00000) [->1]
##1 (reg2[17:1] == 17'h1FFFF) [->1]
##1 c[->1];
endproperty
// BTW, you have too many ()s. Use expr instead (expr)
// You may also want to use the "let" on the path. Thus,
let path=TB.TOP;
path.!c ##1 path.reg2
Ben Cohen
http://www.systemverilog.us/ ben@systemverilog.us
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** SVA Handbook 4th Edition, 2016 ISBN 978-1518681448
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