fix(parser): handle >= and <= operators in (( )) arithmetic commands

[chaliy]

Summary

• Fixed >= and <= operators in (( )) arithmetic commands always evaluating incorrectly
• The tokenizer splits >= into RedirectOut + Word("=") and <= into RedirectIn + Word("="), which caused the (( )) parser to produce malformed expressions
• Added lookahead in the arithmetic command parser to combine these tokens back into >= and <=
• Also handles N>= (e.g., 3>=5) where the tokenizer produces RedirectFd(N) + Word("=...")

Why

(( 3 >= 5 )) always returned exit code 0 (true) because the = was dropped from the expression, producing 3> = 5 which was misinterpreted as an assignment returning 5 (non-zero → true).

Tests

• Added spec tests: arith_cmd_ge_false, arith_cmd_ge_true, arith_cmd_ge_equal, arith_cmd_ge_nospace, arith_cmd_le_nospace
• 100% bash comparison match (1697/1697)

Closes #967