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If and contestant
If the contestant answers a question incorrectly, then all the money won so far is lost, except that the £ 1, 000 and £ 32, 000 prizes are guaranteed: if a player gets a question wrong above these levels, then the prize drops to the previous guaranteed prize.
If the contestant was right, he or she got the square ; if the contestant was wrong, the other contestant got the square, unless that caused the opponent to get three in a row.
If the match ended in a tie, one final question was played with the star of one contestant's choosing ; if the contestant can agree or disagree correctly, he / she won the match, otherwise the match went to the opponent.
If the chosen key started the car, the contestant won it and retired ; otherwise, the contestant returned the next day with that car eliminated should he or she return to the bonus game.
If a contestant failed to match any of the three answers, the bonus round ended.
If no match was made in the Audience Match portion of the Super-Match, the contestant played for $ 500 in the Head-To-Head Match.
If both of the contestant's opponents have the same score, the passing contestant can choose the recipient.
If the contestant chooses to play, a lighted " spinner " begins moving around the board, while the individual squares on the board cycle through a series of items.
If a contestant hits four Whammies, that contestant is eliminated from the game.
If a contestant hits a total of four Whammies at any point in the game, that contestant is automatically eliminated from the game.
If the contestant hit a Whammy, the home player received a $ 500 consolation prize.
If a contestant plays a false hidden immunity idol, the host would toss it into the fire pit.
If the exiled contestant is asked to return after the Tribal Council ( whether they belong to a tribe or not ), they will also be immune from being voted out at the respective Tribal Council.
* If a contestant wants to play the hidden immunity idol, this must be done after the votes have been cast but before they are read.
* If a contestant plays the hidden immunity idol, any votes cast for that contestant will not count, and the person with the next largest number of votes will be eliminated.
If two or more contestants have an equal number of points, then the contestant with the fewer passes is the winner.
If the contestant could not complete the " Truth " portion, there would be " Consequences ", usually a zany and embarrassing stunt.
If a contestant was able to pick all three drawers with money inside before picking the empty drawer, he or she won a bonus prize.

If and becomes
If he becomes chancellor, Dr. Erhart would make few changes.
If we grasp this orientation as a key, our national conduct in all of the events here mentioned becomes intelligible.
If the coating becomes still thicker, a peeling type failure finally can occur.
If we neglect higher than first order terms in the fluctuations, the mass balance equation becomes
If the balance factor becomes 0 then the height of the subtree has decreased by one and the retracing needs to continue.
If the balance factor becomes − 2 or + 2 then the subtree is unbalanced and needs to be rotated to fix it.
If the object point O is infinitely distant, u1 and u2 are to be replaced by h1 and h2, the perpendicular heights of incidence ; the sine condition then becomes sin u ' 1 / h1 = sin u ' 2 / h2.
If the batter gets a base hit, which would have scored the runner anyway, the run now becomes earned.
* If G is a locally compact Hausdorff topological group and μ its Haar measure, then the Banach space L < sup > 1 </ sup >( G ) of all μ-integrable functions on G becomes a Banach algebra under the convolution xy ( g ) = ∫ x ( h ) y ( h < sup >− 1 </ sup > g ) dμ ( h ) for x, y in L < sup > 1 </ sup >( G ).
If we allow only quantifiers, it becomes the Co-NP-complete tautology problem.
If a suitably sized quantum computer capable of running Grover's algorithm reliably becomes available, it would reduce a 128-bit key down to 64-bit security, roughly a DES equivalent.
If a non-working point number placed, bought or laid becomes the new point as the result of a come-out, the bet is usually refunded, or can be moved to another number for free.
If is any unit vector, the projection of the curl of F onto is defined to be the limiting value of a closed line integral in a plane orthogonal to as the path used in the integral becomes infinitesimally close to the point, divided by the area enclosed.
If the core becomes sufficiently dense, electron degeneracy pressure will play a significant part in stabilizing it against gravitational collapse.
If the column load is gradually increased, a condition is reached in which the straight form of equilibrium becomes so-called neutral equilibrium, and a small lateral force will produce a deflection that does not disappear and the column remains in this slightly bent form when the lateral force is removed.
If the 15 minutes of a quarter expire while the ball is live, the quarter is extended until the ball becomes dead.
If the situation worsens it is possible that the defenses cease to function altogether and the individual becomes aware of the incongruence of their situation.
If m is greater than 1, the figure becomes larger ; if m is between 0 and 1, it becomes smaller.
If S is an arbitrary set, then the set S < sup > N </ sup > of all sequences in S becomes a complete metric space if we define the distance between the sequences ( x < sub > n </ sub >) and ( y < sub > n </ sub >) to be, where N is the smallest index for which x < sub > N </ sub > is distinct from y < sub > N </ sub >, or 0 if there is no such index.
If these two features of the language coincide too frequently, they overemphasize each other and the hexameter becomes sing-songy.
If y is a point where the vector field v ( y ) ≠ 0, then there is a change of coordinates for a region around y where the vector field becomes a series of parallel vectors of the same magnitude.
If one alters a Euclidean space so that its inner product becomes negative in one or more directions, then the result is a pseudo-Euclidean space.
If I told you my son's age, then there would no longer be two unknowns ( variables ), and the problem becomes a linear equation with just one variable, that can be solved as described above.
If we allow to incorporate the algebraic sign, this becomes a sum and implies that there is a function of state which is conserved over a complete cycle.

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